Algebra 2

Algebra 2 Practice Test: Practice Test 49

Practice Test 49 for Algebra 2: real questions and explanations from the Varsity Tutors practice-test pool.

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Question 1 of 25

Construct an exponential function from the geometric sequence $3,\ 12,\ 48,\ 192,\ \dots$ where the first term corresponds to $x=1$ (so $f(1)=3$ ). Write $f(x)$ in the form $f(x)=a\cdot b^x$ .

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Question 1

Construct an exponential function from the geometric sequence $3,\ 12,\ 48,\ 192,\ \dots$ where the first term corresponds to $x=1$ (so $f(1)=3$ ). Write $f(x)$ in the form $f(x)=a\cdot b^x$ .

$f(x)=3\cdot 4^x$
$f(x)=12\cdot 4^{x-1}$
$f(x)=\frac{3}{4}\cdot 4^x$ (correct answer)
$f(x)=3\cdot 3^x$

Explanation: This question tests your ability to construct linear or exponential functions from given information like points, tables, graphs, or descriptions of relationships. For exponential functions, find the initial value a (the y-value when x = 0, or work backward if needed) and the growth/decay factor b (divide consecutive y-values: b = y₂/y₁ when x increases by 1). Then write f(x) = a·b^x. To construct a linear function from two points, find the slope m = (y₂ - y₁)/(x₂ - x₁), then find the y-intercept b by substituting one point into y = mx + b and solving for b. Once you have m and b, you've got your function! For this geometric sequence with common ratio 4 and f(1)=3, solve 3 = a*4^1 gives a=3/4, so f(x)=(3/4)*4^x. Choice C correctly constructs the exponential function with coefficient 3/4 and base 4 from the sequence. A distractor like choice A might forget to adjust a for x starting at 1—use f(1) to solve for a! Exponential construction strategy: (1) Find initial value: if you have x = 0 in data, that y is your a; otherwise calculate backward using the pattern, (2) Find base: divide consecutive y-values (with x differing by 1): b = y_{x+1}/y_x—should be constant for exponential, (3) Write f(x) = a·b^x, (4) Verify with all data points. Example: points (0, 100) and (1, 110) and (2, 121) → a = 100, b = 110/100 = 1.1 (check: 121/110 = 1.1 ✓), so f(x) = 100·(1.1)^x. The ratio test both identifies the type and gives you the base!

Question 2

Find the modulus of the complex number $z = 6 - 8i$ . (Recall: $|a+bi|=\sqrt{a^2+b^2}$ , the distance from the origin to $(a,b)$ on the complex plane.)

$|z|=\sqrt{6^2-8^2}=\sqrt{-28}$
$|z|=\sqrt{6^2+8^2}=10$ (correct answer)
$|z|=6+8=14$
$|z|=\sqrt{6^2+8^2}=\sqrt{28}$

Explanation: This question tests your understanding of calculating the modulus (absolute value) of a complex number using the distance formula on the complex plane. The complex plane is a coordinate system where the horizontal axis represents the real part and the vertical axis represents the imaginary part: the complex number a + bi is plotted at point (a, b), just like ordered pairs! The modulus (absolute value) of a + bi equals square root of (a squared + b squared), which is the distance from the origin to point (a, b) using the Pythagorean theorem—the real and imaginary parts form legs of right triangle, modulus is hypotenuse! To find modulus of z = 6 - 8i: (1) Identify real part a = 6 and imaginary part b = -8 (note the negative!). (2) Apply modulus formula: |z| = square root of (6 squared + (-8) squared) = square root of (36 + 64) = square root of 100 = 10. (3) Geometric interpretation: the point (6, -8) is 10 units from origin. This forms a 6-8-10 right triangle, a multiple of the famous 3-4-5 triangle! Choice B correctly calculates |z| = square root of (6 squared + 8 squared) = 10. Choice A incorrectly subtracts the squares instead of adding: square root of (6 squared - 8 squared) would give square root of negative, which isn't real! Modulus uses addition under the square root. Choice C adds the absolute values directly: 6 + 8 = 14, forgetting the squares and square root—distance requires Pythagorean theorem! Choice D makes an arithmetic error: square root of (36 + 64) = square root of 100 = 10, not square root of 28. Modulus calculation checklist: (1) Square both parts (signs don't matter after squaring). (2) Add the squares. (3) Take square root. The result is always non-negative since it represents distance!

Question 3

A radioactive substance has a half-life of 5 years. If there are initially 800 grams of the substance, which function models the amount remaining after $t$ years?

$A(t) = 800(0.5)^t$
$A(t) = 800(0.5)^{t/5}$ (correct answer)
$A(t) = 800(0.1)^{t/5}$
$A(t) = 800 - 160t$

Explanation: For exponential decay with half-life, $A(t) = A_0(0.5)^{t/h}$ where $h$ is the half-life. With $A_0 = 800$ and $h = 5$ , we get $A(t) = 800(0.5)^{t/5}$ . This gives $A(5) = 800(0.5)^1 = 400$ , which is correct for one half-life. Choice A would halve the substance every year, not every 5 years. Choice C uses the wrong base. Choice D represents linear decay, not exponential.

Question 4

Determine the inverse function for $f(x)=3x-7$ .

$f^{-1}(x)=3x+7$
$f^{-1}(x)=\dfrac{x-7}{3}$
$f^{-1}(x)=\dfrac{x+7}{3}$ (correct answer)
$f^{-1}(x)=\dfrac{3}{x-7}$

Explanation: This question tests your ability to find inverse functions—functions that undo what the original function does, reversing the input-output relationship. An inverse function f⁻¹(x) reverses f(x): if f takes a to b, then f⁻¹ takes b back to a. To find the inverse algebraically, use the swap-and-solve method: (1) write y = f(x), (2) swap x and y (this reverses the input-output roles), (3) solve for y, (4) the result is y = f⁻¹(x). For f(x) = 3x - 7, we write y = 3x - 7, swap to get x = 3y - 7, then solve for y: add 7 to both sides to get x + 7 = 3y, then divide by 3 to get y = (x+7)/3, so f⁻¹(x) = (x+7)/3. Choice C correctly finds f⁻¹(x) = (x+7)/3 by swapping and solving properly—it undoes 'multiply by 3 then subtract 7' with 'add 7 then divide by 3.' Choice B incorrectly subtracts 7 instead of adding 7, while Choice D confuses inverse with reciprocal (3/(x-7) is NOT the inverse!). The swap-and-solve recipe ensures you reverse operations in the correct order: if f does 'multiply by 3, then subtract 7,' the inverse must 'add 7, then divide by 3.' Inverse thinking helps predict the answer before computing!

Question 5

A student is choosing how many hours to spend tutoring in math and science this week. Let $x$ be hours of math tutoring and $y$ be hours of science tutoring.

Constraints:

Total tutoring time is at most 12 hours: $x+y\le 12$ .
The student wants at least twice as much math as science: $x\ge 2y$ .
At least 3 hours of science: $y\ge 3$ .
Non-negativity: $x\ge 0$ , $y\ge 0$ .

Determine which solutions are viable: $(6,3)$ , $(5,3)$ , $(8,4)$ .

Only $(8,4)$ is viable.
All three points are viable.
$(6,3)$ and $(8,4)$ are viable. (correct answer)
Only $(6,3)$ is viable.

Explanation: This question tests your ability to translate real-world limitations into mathematical constraints (equations and inequalities) and determine whether potential solutions are viable—meaning they satisfy all constraints and make sense in context. Checking viability is systematic: (1) write out each constraint, (2) substitute the proposed solution into each one, (3) verify each is satisfied (inequality holds, equation balances), (4) check context reasonableness (non-negative? integers if needed?). If everything passes, it's viable. If anything fails, it's nonviable—and you should identify WHICH constraint was violated. Complete checking means checking ALL constraints, not just some! For (6,3): 6+3=9≤12, 6≥2*3=6, 3≥3 (all ✓); for (8,4): 8+4=12≤12, 8≥8, 4≥3 (all ✓); but (5,3): 5<6 violates x≥2y. Choice C correctly determines viability with proper reasoning for both points. A distractor like choice A might miss checking (8,4) fully, but gently substitute into each constraint for all points to confirm. Constraint identification from context: (1) list every limitation mentioned ('budget $X,' 'time ≤ Y hours,' 'need ≥ Z units'), (2) translate using key phrases: 'at most' → ≤, 'at least' → ≥, 'exactly' → =, 'more than' → >, 'less than' → <, (3) don't forget implicit constraints like x ≥ 0, y ≥ 0 (can't be negative) or x, y integers (if discrete), (4) write the complete system. Missing even one constraint can make you accept infeasible solutions! The viability checklist: Make a table with one row per constraint. For each constraint, substitute your point and mark whether it's satisfied (✓) or violated (✗). If all checks pass AND context is reasonable, mark VIABLE. If even one ✗ appears OR context is violated (like negative values or fractional items), mark NONVIABLE and note which constraint failed. This organized approach prevents missing checks and makes your reasoning clear. Thorough constraint checking is what separates good modeling from sloppy work!

Question 6

The function $p(x)$ satisfies $p(-x) = -p(x)$ for all $x$ . If $h(x) = p(x-3) + 2$ , which property does $h(x)$ possess?

$h(x)$ is even: $h(-x) = h(x)$
$h(x)$ is odd: $h(-x) = -h(x)$
$h(x)$ is symmetric about the point $(3, 2)$ (correct answer)
$h(x)$ is symmetric about the line $x = 3$

Explanation: Since $p(-x) = -p(x)$ , $p(x)$ is odd and symmetric about the origin. For $h(x) = p(x-3) + 2$ , this shifts the graph right 3 units and up 2 units. An odd function shifted right and up becomes symmetric about the point where the origin moves to, which is $(3, 2)$ . To verify: $h(3+t) + h(3-t) = p(t) + 2 + p(-t) + 2 = p(t) - p(t) + 4 = 4 = 2 \cdot 2$ , confirming point symmetry about $(3,2)$ . Choice A is wrong because horizontal shifts destroy evenness. Choice B is wrong because vertical shifts destroy oddness. Choice D describes line symmetry, not point symmetry.

Question 7

Write the system as $AX=B$ (use variable order $x, y, z$ ): $\begin{cases} -x + 2y + z = 3 \\ 4x - y = 0 \\ 2x + 5y - 3z = -7 \end{cases}$

$\begin{bmatrix}-1&2&1\\4&-1&0\\2&5&-3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}3\\-7\\0\end{bmatrix}$
$\begin{bmatrix}-1&2&1\\4&-1&1\\2&5&-3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}3\\0\\-7\end{bmatrix}$
$\begin{bmatrix}-1&2&1\\4&-1&0\\2&5&-3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}3\\0\\-7\end{bmatrix}$ (correct answer)
$\begin{bmatrix}-1&4&2\\2&-1&5\\1&0&-3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}3\\0\\-7\end{bmatrix}$

Explanation: This question tests your ability to represent a system of linear equations as a single matrix equation in the form $AX = B$ , where $A$ is the coefficient matrix, $X$ is the variable vector, and $B$ is the constant vector, including zeros for missing variables. Matrix form $AX = B$ is a compact way to write entire systems: the coefficient matrix $A$ contains all the coefficients from the left sides of equations (organized by rows for equations and columns for variables), the variable vector $X$ lists the unknowns as a column, and the constant vector $B$ lists the right-hand side values. When you multiply matrix $A$ times vector $X$ , you get the left sides of all equations, which equals vector $B$ (the right sides). For example, the system $2x + 3y = 7$ and $x - y = 1$ becomes $\begin{bmatrix} 2 & 3 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 7 \\ 1 \end{bmatrix}$ . Matrix multiplication recovers the original equations! To convert the system $-x + 2y + z = 3$ , $4x - y = 0$ , $2x + 5y - 3z = -7$ to matrix form (order $x, y, z$ ): $A = \begin{bmatrix} -1 & 2 & 1 \\ 4 & -1 & 0 \\ 2 & 5 & -3 \end{bmatrix}$ , $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ , $B = \begin{bmatrix} 3 \\ 0 \\ -7 \end{bmatrix}$ —note the 0 for missing $z$ in the second equation. Choice A correctly includes the zero for $z$ in row 2, with all coefficients and constants in proper order. Choice D transposes parts of $A$ , like putting rows into columns, which scrambles the coefficients—stick to rows as equations and columns as variables! Matrix equation construction recipe: (1) Label equations (first, second, third, etc.) and variables ( $x, y, z$ in consistent order), (2) Build matrix $A$ : make rows from equations (row i = coefficients from equation i), make columns from variables (column j = coefficient of variable j in each equation), include 0 for missing variables, (3) Write $X$ as column vector with variables in same order as $A$ 's columns, (4) Write $B$ as column vector with constants in same order as equations, (5) Combine as $AX = B$ . Example: $x + 2y = 5$ and $3x - y = 7$ gives $A = \begin{bmatrix} 1 & 2 \\ 3 & -1 \end{bmatrix}$ , $X = \begin{bmatrix} x \\ y \end{bmatrix}$ , $B = \begin{bmatrix} 5 \\ 7 \end{bmatrix}$ . Systematic! To verify your matrix equation represents the system correctly, perform the matrix multiplication $AX$ mentally or on paper: the first row of $A$ times $X$ should give the left side of equation 1, the second row times $X$ should give equation 2's left side, etc. If $\begin{bmatrix} 2 & 3 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}$ gives $\begin{bmatrix} 2x + 3y \\ x - y \end{bmatrix}$ , and $B = \begin{bmatrix} 7 \\ 1 \end{bmatrix}$ , then matrix equation represents $2x + 3y = 7$ and $x - y = 1$ . Matrix multiplication recovers the system—this verification prevents errors!

Question 8

A number and its reciprocal have a sum of $\frac{10}{3}$ . Write and solve an equation to find the number.

Let $x$ = the number, with $x\ne 0$ .

Set up $x+\frac{1}{x}=\frac{10}{3}$ . Then $3x^2-10x+3=0$ , so $x=3$ or $x=\frac{1}{3}$ . Both satisfy the context. (correct answer)
Set up $x-\frac{1}{x}=\frac{10}{3}$ . Then $x=3$ only.
Set up $\frac{1}{x}=\frac{10}{3}x$ . Then $x=\pm\sqrt{\frac{3}{10}}$ .
Set up $x+\frac{1}{x}=\frac{10}{3}$ . Then $x=\frac{10}{3}$ only.

Explanation: This question tests your ability to translate complex real-world situations into mathematical equations (linear, quadratic, rational, or exponential) and solve them to answer practical questions. Problems involving a number and its reciprocal lead to rational equations: if the number is x, its reciprocal is 1/x, so their sum gives x + 1/x = 10/3. Multiplying through by x: x² + 1 = 10x/3, then 3x² + 3 = 10x, so 3x² - 10x + 3 = 0. Using the quadratic formula or factoring: (3x - 1)(x - 3) = 0, giving x = 1/3 or x = 3, and both are valid since both have reciprocals. Choice A correctly sets up the equation as number + reciprocal = sum (x + 1/x = 10/3) and solves to get x = 3 or x = 1/3, noting both satisfy the context. Choice B uses subtraction instead of addition, C sets up an incorrect equation 1/x = 10x/3, and D claims only x = 10/3 is the solution without solving the quadratic. Context-to-equation type matching: (1) Constant rates and simple relationships → linear, (2) Area, projectile motion, optimization → quadratic, (3) Combined work rates, mixture concentrations, reciprocal relationships → rational, (4) Percent growth/decay over time → exponential. The context language tells you which type: 'per unit' suggests linear, 'area' suggests quadratic, 'together complete' suggests rational (adding reciprocals), 'percent per year' suggests exponential. Identify the type, then use the appropriate solving method! The reality check is essential: after solving, ask: (1) Does the answer satisfy the original equation? (substitute back), (2) Does it make sense in the real world? (no negative quantities, no fractional discrete items), (3) Have I answered what was asked? (find time, not rate; find width, not area). For rational equations, check that denominators aren't zero. For exponential, verify the value is reasonable for the time scale. This three-part check catches most errors and ensures your math answer is also a real-world answer!

Question 9

A tank is being filled by two hoses. Hose A can fill the tank in 6 hours, and Hose B can fill the tank in 8 hours. Write and solve an equation to find how long it takes to fill the tank if both hoses run together.

Let $t$ = the number of hours to fill the tank together.

Set up $\frac{1}{6}+\frac{1}{8}=t$ . Then $t=\frac{7}{24}$ hour. The tank fills in $\frac{7}{24}$ hour.
Set up $\frac{1}{6}+\frac{1}{8}=\frac{1}{t}$ . Then $t=\frac{24}{7}$ hours. The tank fills in $\frac{24}{7}$ hours. (correct answer)
Set up $\frac{6}{t}+\frac{8}{t}=1$ . Then $t=14$ hours. The tank fills in 14 hours.
Set up $\frac{1}{6}-\frac{1}{8}=\frac{1}{t}$ . Then $t=24$ hours. The tank fills in 24 hours.

Explanation: This question tests your ability to translate complex real-world situations into mathematical equations (linear, quadratic, rational, or exponential) and solve them to answer practical questions. Work-rate problems lead to rational equations: if one worker completes a job in time t₁ and another in t₂, their combined rate is 1/t₁ + 1/t₂ (adding rates), which equals 1/t_combined. The reciprocals represent 'fraction of job per hour,' and adding these fractions gives the combined rate. Solving these rational equations requires finding LCD and often produces fractional time answers that make sense: 3.4 hours = 3 hours 24 minutes. Hose A fills 1/6 of the tank per hour, Hose B fills 1/8 per hour, so together they fill 1/6 + 1/8 = 4/24 + 3/24 = 7/24 of the tank per hour. To find the time to fill one whole tank, we set up 1/6 + 1/8 = 1/t, which gives 7/24 = 1/t, so t = 24/7 hours (about 3.43 hours). Choice B correctly sets up the equation as rate₁ + rate₂ = combined rate, yielding t = 24/7 hours. Choice A incorrectly adds rates to get time directly (1/6 + 1/8 = t), which would mean the tank fills in 7/24 of an hour (about 17.5 minutes) - far too fast since even the faster hose alone takes 6 hours! Context-to-equation type matching: (1) Constant rates and simple relationships → linear, (2) Area, projectile motion, optimization → quadratic, (3) Combined work rates, mixture concentrations, reciprocal relationships → rational, (4) Percent growth/decay over time → exponential. The context language tells you which type: 'per unit' suggests linear, 'area' suggests quadratic, 'together complete' suggests rational (adding reciprocals), 'percent per year' suggests exponential. Identify the type, then use the appropriate solving method! The reality check is essential: after solving, ask: (1) Does the answer satisfy the original equation? (substitute back), (2) Does it make sense in the real world? (no negative quantities, no fractional discrete items), (3) Have I answered what was asked? (find time, not rate; find width, not area). For rational equations, check that denominators aren't zero. For exponential, verify the value is reasonable for the time scale. This three-part check catches most errors and ensures your math answer is also a real-world answer!

Question 10

A profit function is $P(x)=x^2-8x+15$ , where zeros represent break-even points. Factor $P(x)$ and find the break-even x-values.

$P(x)=(x+3)(x+5)$ ; break-even x-values: $x=-3,-5$
$P(x)=(x-1)(x-15)$ ; break-even x-values: $x=1,15$
$P(x)=(x-3)(x-5)$ ; break-even x-values: $x=3,5$ (correct answer)
$P(x)=(x-8)(x+7)$ ; break-even x-values: $x=8,-7$

Explanation: This question tests your ability to factor quadratic expressions and use the factored form to identify zeros (x-intercepts) of the function—essential for graphing and solving quadratic equations; here, zeros are break-even points. Factoring a quadratic into form f(x) = a(x - r)(x - s) reveals the zeros immediately: set the factored expression equal to zero and use the zero product property (if a product equals zero, at least one factor must equal zero). Setting (x - r) = 0 gives x = r, and setting (x - s) = 0 gives x = s, so the zeros are r and s—these are the x-intercepts of the parabola at points (r, 0) and (s, 0) where the graph crosses the x-axis. For P(x) = x² - 8x + 15, find numbers multiplying to 15 adding to -8: -3 and -5, so (x - 3)(x - 5); break-even at x = 3 and x = 5—positive zeros make sense for profits! Choice A correctly factors and identifies the break-even points. Choice B flips signs, giving positive factors that expand to x² + 8x + 15—signs are key, opposite for negative sum! Apply this to real-world models: factors reveal intercepts directly; practice sign relationships to avoid common errors—you're building strong skills!

Question 11

Describe the end behavior of the exponential function $h(x)=5(0.7)^x$ . (Relate to the parent function $y=(0.7)^x$ .)

As $x\to\infty$ , $h(x)\to\infty$ ; as $x\to-\infty$ , $h(x)\to 0$
As $x\to\infty$ , $h(x)\to 0$ ; as $x\to-\infty$ , $h(x)\to\infty$ (correct answer)
As $x\to\infty$ , $h(x)\to 5$ ; as $x\to-\infty$ , $h(x)\to 0$
As $x\to\infty$ , $h(x)\to -\infty$ ; as $x\to-\infty$ , $h(x)\to\infty$

Explanation: This question tests your ability to analyze exponential functions by identifying their end behavior based on the base value. Exponential functions f(x) = ab^x have distinctive features: they have a y-intercept at (0, a) because b⁰ = 1, they never cross the x-axis (no x-intercepts for basic form), and they have a horizontal asymptote at y = 0 (the x-axis) that the graph approaches but never touches. End behavior depends on the base: if b > 1, it's exponential growth (left end approaches 0, right end goes to ∞); if 0 < b < 1, it's exponential decay (left end goes to ∞, right end approaches 0). Transformations like f(x) = ab^x + k shift the horizontal asymptote to y = k! For h(x) = 5(0.7)^x, the base is 0.7, which is between 0 and 1, indicating exponential decay. As x → ∞, the term (0.7)^x approaches 0 (since we're multiplying 0.7 by itself more and more times), so h(x) → 5·0 = 0. As x → -∞, the term (0.7)^x becomes (0.7)^(-|x|) = 1/(0.7)^|x|, which approaches ∞, so h(x) → ∞. Choice B correctly states: as x → ∞, h(x) → 0; as x → -∞, h(x) → ∞. Choice A reverses the behavior (that's for growth functions with b > 1), while Choice C incorrectly suggests the function approaches 5. Exponential end behavior strategy: Remember the base determines everything! If 0 < b < 1 (decay), the function decreases as x increases, approaching 0 on the right and ∞ on the left. If b > 1 (growth), it's the opposite. The coefficient a only affects the y-intercept and vertical stretch, not the end behavior pattern!

Question 12

In the expression $2x^3 - 5x^2 + x - 8$ , how many terms are there, and what are they? (Remember: terms are separated by $+$ or $-$ , and the sign belongs to the term.)

3 terms: $2x^3 - 5x^2$ , $x$ , $-8$
6 terms: $2$ , $x^3$ , $-5$ , $x^2$ , $x$ , $-8$
4 terms: $2x^3$ , $5x^2$ , $x$ , $8$
4 terms: $2x^3$ , $-5x^2$ , $x$ , $-8$ (correct answer)

Explanation: This question tests your understanding of the structure of algebraic expressions—specifically, how to identify terms, which are the parts separated by addition or subtraction signs. Terms are the pieces of an expression separated by plus or minus signs; for example, in 3x² - 5x + 7, there are three terms: 3x², -5x, and +7, with the sign belonging to each term. In the expression 2x³ - 5x² + x - 8, the terms are 2x³, -5x², +x, and -8, making a total of four terms—remember to include the sign with each term after the first. Choice A correctly identifies the four terms by properly separating them at the + and - signs and including their signs. A tempting distractor like Choice D confuses terms with factors by breaking each term into its multiplicative parts, such as listing 2 and x³ separately, but remember that terms are additive parts, not multiplicative ones. To count terms reliably, look for the number of + and - signs that separate the expression and add one—for this expression, there are three separating signs (-, +, -), so 3 + 1 = 4 terms. Always list terms with their full signs to avoid confusion, and keep practicing this to build confidence in analyzing polynomials—you're doing great!

Question 13

The function $g(x)=-2x^3+5x$ models the net water flow rate (in liters/minute) into a tank as a function of valve setting $x$ . Describe the end behavior of $g$ .

As $x\to\infty$ , $g(x)\to\infty$ and as $x\to-\infty$ , $g(x)\to-\infty$ .
As $x\to\infty$ , $g(x)\to-\infty$ and as $x\to-\infty$ , $g(x)\to\infty$ . (correct answer)
As $x\to\infty$ , $g(x)\to 5$ and as $x\to-\infty$ , $g(x)\to 5$ .
As $x\to\infty$ , $g(x)\to\infty$ and as $x\to-\infty$ , $g(x)\to\infty$ .

Explanation: This question tests your ability to identify and interpret key features of functions—like intercepts, where they increase or decrease, maximum and minimum values, end behavior, and (for some functions) periodicity—from graphs, tables, or formulas. Key features tell the complete story of a function: intercepts show where it crosses the axes (starting value or zeros), increasing/decreasing intervals show where it's rising or falling, extrema show peaks and valleys (best/worst outcomes), and end behavior describes long-term trends. For g(x) = -2x³ + 5x, the leading term -2x³ dominates for large |x|, and since it has odd degree with negative coefficient, as x→∞, g(x)→-∞ and as x→-∞, g(x)→∞. Choice B correctly describes this end behavior, showing that for very large positive valve settings, water flows out rapidly (negative flow), while for large negative settings, water flows in rapidly. Choice A reverses these directions, ignoring the negative coefficient on the leading term. Sketching from features: end behavior is determined by the leading term—for polynomials, look at the degree (odd/even) and leading coefficient sign (positive/negative) to quickly determine where the graph heads!

Question 14

A quantity $Q$ follows the pattern $Q_{n+1}=0.92\,Q_n$ each week.

What percent does the quantity change each week, and is it growth or decay?

8% growth per week because $0.92+0.08=1$
0.08% decay per week because $1-0.92=0.08$
8% decay per week because $0.92=1-0.08$ (correct answer)
92% decay per week because the factor is $0.92$

Explanation: This question tests your ability to recognize exponential growth or decay—situations where a quantity changes by a constant percent (not constant amount) per time interval. Constant percent growth means multiplying by the same factor greater than 1 each interval: if something grows 5% per year, each year's value is 105% of the previous (multiply by 1.05). Constant percent decay means multiplying by a factor between 0 and 1: if something decreases 15% per year, each year retains 85% of previous (multiply by 0.85). The key test: calculate ratios of consecutive values. If ratios are constant, it's exponential with that ratio as the growth/decay factor! The pattern Q_{n+1} = 0.92Q_n means each week's quantity is 92% of the previous week (multiply by 0.92). Since 0.92 < 1, this is decay. To find the percent decay: 0.92 = 1 - 0.08, so the quantity decreases by 8% each week. Choice A correctly identifies 8% decay per week because 0.92 = 1 - 0.08. Choice B incorrectly interprets 0.92 as meaning 92% decay, when 0.92 actually means retaining 92% (losing only 8%). The ratio test for exponential: (1) from table, divide consecutive y-values: y₂/y₁, y₃/y₂, y₄/y₃, (2) if all equal → exponential with that ratio as base b, (3) if b > 1 → growth; if 0 < b < 1 → decay, (4) calculate percent rate: r = b - 1, convert to percent. Example: ratios all 1.06 → base 1.06 → growth → rate = 1.06 - 1 = 0.06 = 6% per interval. This systematic check identifies exponential patterns reliably! Don't confuse exponential with linear: linear adds the same amount each time (constant differences like +50, +50, +50), exponential multiplies by same factor (constant ratios like ×1.1, ×1.1, ×1.1). Both are patterns of regular change, but different mechanisms!

Question 15

Use Pascal's Triangle (row 5: $1,5,10,10,5,1$ ) to expand $(x+y)^5.$

$x^5+5x^4y+10x^3y^2+5x^2y^3+10xy^4+y^5$
$x^5+4x^4y+6x^3y^2+4x^2y^3+xy^4+y^5$
$x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$ (correct answer)
$x^5+5x^4y+10x^2y^3+10x^3y^2+5xy^4+y^5$

Explanation: This question tests your understanding of the Binomial Theorem—a formula for expanding $(x + y)^n$ using coefficients from Pascal's Triangle, which saves enormous time compared to multiplying out by hand! The Binomial Theorem says $(x + y)^n$ expands to a sum of n+1 terms where coefficients come from row n of Pascal's Triangle, x-powers decrease from n to 0, and y-powers increase from 0 to n. For $(x + y)^5$ , we use row 5 of Pascal's Triangle ( $1, 5, 10, 10, 5, 1$ ) to get: $1 \cdot x^5 + 5 \cdot x^4 y + 10 \cdot x^3 y^2 + 10 \cdot x^2 y^3 + 5 \cdot x y^4 + 1 \cdot y^5$ . Choice A correctly shows this expansion with all six terms having the right coefficients from Pascal's Triangle and powers that decrease for x ( $5 \to 0$ ) while increasing for y ( $0 \to 5$ ). Choice B incorrectly swaps the middle coefficients (has 5 and 10 instead of 10 and 10), while Choice C uses row 4 coefficients instead of row 5. The expansion recipe using Pascal's Triangle: (1) Identify $n = 5$ , (2) Use row 5: $1, 5, 10, 10, 5, 1$ , (3) Create 6 terms with decreasing x-powers and increasing y-powers, (4) Write it out systematically—the pattern is beautiful and reliable!

Question 16

From the graph, determine if the rate of change is constant.

The curve passes through $(0,0)$ , $(1,1)$ , $(2,4)$ , and $(3,9)$ and is smooth and curved upward (not a straight line).

Yes; because the points increase, the rate of change is constant.
No; the graph is curved, so the slope changes and the rate of change is non-constant (nonlinear). (correct answer)
Yes; because $\Delta y$ is always 3 when $\Delta x=1$ .
No; because the $y$ -intercept is 0, the rate of change must be zero.

Explanation: This question tests your ability to recognize when a relationship has constant rate of change—the defining characteristic of linear functions. Constant rate of change means that for every unit increase in x, y changes by the same amount every time: if Δy/Δx = 5 for one interval and also 5 for every other equal interval, the rate is constant at 5. This constant rate is exactly what makes a function linear (y = mx + b where m is that constant rate). Only linear functions have this property—quadratics, exponentials, and other nonlinear functions have rates that vary at different x-values! The graph is curved upward through (0,0), (1,1), (2,4), (3,9), so slopes vary: e.g., from (0,0) to (1,1) rate=1, from (1,1) to (2,4) rate=3, from (2,4) to (3,9) rate=5, non-constant. Choice B correctly identifies the non-constant rate because the curve means changing slope, indicating nonlinear. A distractor like choice A might assume any increasing graph has constant rate, but curvature shows varying rates—look for straightness to confirm constancy. The three-method constant rate test: METHOD 1 (from table): Calculate Δy/Δx for each pair of consecutive points with equal Δx. All equal? Constant rate. Vary? Non-constant. METHOD 2 (from graph): Is it a straight line? Yes = constant rate. Curved? Non-constant. METHOD 3 (from formula): Is it y = mx + b form? Yes = constant rate m. Any other form (x², b^x, etc.)? Non-constant. Pick the method matching your representation! Don't confuse constant RATE with constant RATIO: constant rate (Δy/Δx equal) characterizes linear functions, constant ratio (y₂/y₁ equal) characterizes exponential functions. Check BOTH in a table: if differences are 3, 3, 3 → linear with rate 3. If ratios are 2, 2, 2 → exponential with base 2. If neither constant → some other type. Knowing which pattern to look for prevents confusing linear with exponential growth!

Question 17

A water tank is being filled at a constant rate. The volume of water (in gallons) after $t$ minutes is $V(t)=25t+10.$ The tank is monitored from $t=0$ minutes until $t=18$ minutes. What domain makes sense for this situation, and should the graph be discrete points or continuous?

Discrete points, $t\in\{0,1,2,\dots,18\}$
Discrete points, all integers $t\ge 0$
Continuous, $t\in[0,18]$ (correct answer)
Continuous, $t\in(0,18)$

Explanation: This question tests your understanding of how a function's domain relates both to its graph (which x-values have points) and to the real-world context (which values make sense practically). Discrete vs continuous domains depend on what you're measuring: if counting distinct objects (people, tickets, items sold), the domain is discrete—only integers work, graph as separate dots. If measuring continuous quantities (time, distance, temperature, money as a continuous quantity), the domain is continuous—any value in an interval works, graph as connected line/curve. The physical nature of the quantity determines which! You can have 2.7 hours but not 2.7 people. Here, t is time in minutes for filling a tank at a constant rate, which is continuous as time and volume can take any value in the interval, monitored from 0 to 18 inclusive, so the graph should be a continuous line. Choice B correctly identifies it as continuous with $t \in [0,18]$ based on the ongoing filling process. Choice A fails by assuming discrete points, but time doesn't jump in whole minutes—fractions are possible. Discrete vs continuous quick-check: ask 'can there be in-between values?' If you're modeling number of students in a classroom, going from 20 to 21 students happens instantly (no 20.5 students exist)—discrete! If you're modeling temperature change from 20°C to 21°C, every value in between occurs—continuous! Context tells you: countable/distinct objects → discrete (dots on graph), measurable/varying quantities → continuous (line/curve on graph). This determines how you graph and what domain notation to use!

Question 18

Complete the square: Rewrite $x^2+y^2-4x+8y=5$ in standard form and identify $r^2$ .

$(x-2)^2+(y+4)^2=25$ (correct answer)
$(x+2)^2+(y-4)^2=25$
$(x-2)^2+(y+4)^2=5$
$(x-4)^2+(y+8)^2=5$

Explanation: This question tests your understanding of how to derive circle equations using the Pythagorean Theorem (distance formula) or find a circle's center and radius by completing the square. When a circle equation is in general form x² + y² + Dx + Ey + F = 0, we complete the square in both x and y to reveal the center and radius: group x-terms (x² + Dx) and complete the square by adding (D/2)², do the same for y-terms with (E/2)², then rearrange to get (x + D/2)² + (y + E/2)² = (D/2)² + (E/2)² - F. The center is (-D/2, -E/2) and radius is √[(D/2)² + (E/2)² - F]. It's completing the square twice, once for each variable! For x² + y² - 4x + 8y = 5, group and complete: (x² - 4x) + (y² + 8y) = 5, add 4 and 16 to both sides for (x - 2)² + (y + 4)² = 25. Choice A correctly completes the square to (x - 2)² + (y + 4)² = 25. Choice C uses the correct form but forgets to square the radius properly, using 5 instead of 25. Completing the square for circles: (1) Group x-terms together and y-terms together: (x² + Dx) + (y² + Ey) = -F, (2) Complete square in x by adding (D/2)² to both sides, (3) Complete square in y by adding (E/2)² to both sides, (4) Factor: (x + D/2)² + (y + E/2)² = (D/2)² + (E/2)² - F, (5) Read center as (-D/2, -E/2) and radius as √[right side]. Example: x² + y² + 6x - 8y = 0 → (x + 3)² + (y - 4)² = 25, so center (-3, 4), radius 5.

Question 19

A point $(a,b)$ on the complex plane represents the complex number $a+bi$ .

Which complex number corresponds to the point $(-2,-7)$ ?

$-2+7i$
$-7-2i$
$-2-7i$ (correct answer)
$2-7i$

Explanation: This question tests your understanding of the correspondence between points on the complex plane and complex numbers. The complex plane is a coordinate system where the horizontal axis represents the real part and the vertical axis represents the imaginary part: the complex number a + bi is plotted at point (a, b), just like ordered pairs! For example, 3 + 2i goes at (3, 2), and -1 - 4i goes at (-1, -4). The first coordinate is always the real part, the second is the imaginary part. For the point (-2, -7): (1) The x-coordinate -2 is the real part. (2) The y-coordinate -7 is the imaginary part. (3) The complex number is -2 + (-7)i = -2 - 7i. Remember that negative imaginary parts are written with subtraction: we write -2 - 7i, not -2 + (-7)i, for clarity! Choice C correctly identifies the complex number as -2 - 7i. Choice A incorrectly makes the imaginary part positive, giving -2 + 7i, which would correspond to point (-2, 7), not (-2, -7). Choice B swaps the real and imaginary parts, giving -7 - 2i for point (-7, -2). Choice D has the wrong sign on the real part, giving 2 - 7i for point (2, -7). Point to complex number recipe: (1) Point (a, b) corresponds to complex number a + bi. (2) First coordinate → real part, second coordinate → imaginary part. (3) If b is negative, write as a - |b|i for standard form. (4) Check: complex number a + bi plots back at point (a, b). The correspondence is one-to-one—every point has exactly one complex number!

Question 20

All points $(x,y)$ that are 6 units from the origin form a circle.

Write an equation in two variables for this circle.
Choose the best graph setup (labels and scale) to draw it accurately.

Which choice is correct?

Equation: $x+y=36$ ; Graph: x-axis $x$ (units), y-axis $y$ (units); equal scales on both axes, from $-8$ to $8$ by 1
Equation: $x^2+y^2=6$ ; Graph: x-axis $x$ (units), y-axis $y$ (units); equal scales on both axes, from $-8$ to $8$ by 1
Equation: $x^2+y^2=36$ ; Graph: x-axis $x$ (units), y-axis $y$ (units); equal scales on both axes, from $-8$ to $8$ by 1 (correct answer)
Equation: $(x-6)^2+(y-6)^2=36$ ; Graph: x-axis $x$ (units), y-axis $y$ (units); x-scale $-8$ to $8$ by 2, y-scale $-20$ to $20$ by 5

Explanation: This question tests your ability to translate real-world relationships into mathematical equations with two or more variables and set up graphs with appropriate labels and scales to visualize these relationships. Creating equations from contexts requires identifying: (1) which quantities vary (your variables), (2) which depends on which (independent vs dependent), (3) the mathematical relationship connecting them (linear rate, quadratic area, exponential growth, etc.). For 'cost is $40 per hour plus$ 15 per item,' the dependent quantity is cost (C), independent quantities are hours (h) and items (n), and the relationship is additive with rates: C = 40h + 15n. The equation structure mirrors the context structure! Distance 6 from origin gives x² + y² = 36, a circle equation; graph with x-axis x (units), y-axis y (units), equal scales -8 to 8 by 1 to show symmetry without distortion. Choice A correctly creates the equation x²+y²=36 and sets up the graph with equal scales covering beyond radius 6. Choice B uses =6 instead of 36, forgetting to square the radius—remember, the formula is x² + y² = r², so square both sides of distance=6. Equation creation framework: (1) Define your variables clearly—'Let x = horizontal distance (units), y = vertical distance (units)'—being specific prevents confusion, (2) Identify the mathematical structure from context language: 'per' means multiply (rate), 'plus' means add, 'times' or 'product' means multiply, 'percent' means exponential, (3) Build the equation piece by piece matching each phrase in the context, (4) Verify with a test value: does your equation give sensible output for a reasonable input? This catches setup errors before solving! Graph scale decision process: (1) Find your data range—what are the minimum and maximum values you need to show for each axis? (2) Divide that range by 5-10 to get interval size, (3) Round to a 'nice' number: use 1, 2, 5, 10, 20, 50, 100, etc. (not 7 or 13!), (4) Mark intervals starting at 0 (or other logical point). Example: data from 0 to 80 → range is 80, divided by 8 gives 10, so mark every 10: 0, 10, 20, ..., 80. Clean, readable, shows all data!

Question 21

Graph $f(x)=\frac{(x-4)(x+1)}{(x-2)(x+3)}$ by identifying its zeros, vertical asymptotes, and horizontal asymptote.

Zeros: $x=4,-1$ ; VAs: $x=2,-3$ ; HA: $y=1$ (correct answer)
Zeros: $x=2,-3$ ; VAs: $x=4,-1$ ; HA: $y=1$
Zeros: $x=4,-1$ ; VAs: $x=2,-3$ ; HA: $y=0$
Zeros: $x=4,1$ ; VAs: $x=2,-3$ ; HA: $y=1$

Explanation: This question tests your ability to graph rational functions by finding zeros (from the numerator), vertical asymptotes (from the denominator), and horizontal or oblique asymptotes (from comparing degrees and using end behavior). Rational functions f(x) = p(x)/q(x) have distinctive features: zeros where p(x) = 0 (numerator equals zero, these are x-intercepts), vertical asymptotes where q(x) = 0 (denominator equals zero, graph shoots to ±∞), and horizontal or oblique asymptotes describing end behavior. The key is: numerator gives zeros, denominator gives vertical asymptotes. IMPORTANT: if numerator and denominator share a factor (like both have (x - 2)), that creates a hole (removable discontinuity) at x = 2, not a zero or asymptote—the common factor cancels! For f(x) = (x-4)(x+1)/(x-2)(x+3), zeros occur when numerator = 0: (x-4)(x+1) = 0 gives x = 4 and x = -1. Vertical asymptotes occur when denominator = 0: (x-2)(x+3) = 0 gives x = 2 and x = -3. Since both numerator and denominator have degree 2 (equal degrees), the horizontal asymptote is y = 1/1 = 1. Choice A correctly identifies zeros at x = 4, -1; vertical asymptotes at x = 2, -3; and horizontal asymptote at y = 1. Choice B incorrectly swaps zeros and VAs—a common mistake when students confuse which polynomial gives which feature. The rational function feature-finding roadmap: (1) ZEROS: set numerator = 0, solve (these are x-intercepts), (2) VERTICAL ASYMPTOTES: set denominator = 0, solve (but check for common factors with numerator—if common, it's a hole, not VA!), (3) HORIZONTAL ASYMPTOTE: compare degrees: deg(num) < deg(den) → y = 0; degrees equal → y = ratio of leading coefficients; deg(num) > deg(den) → no HA. (4) OBLIQUE ASYMPTOTE: if deg(num) = deg(den) + 1, divide to find it. Follow these steps systematically! Always remember the fundamental rule: numerator determines zeros (where the graph crosses the x-axis), denominator determines vertical asymptotes (where the graph has vertical barriers).

Question 22

The amount of a medication remaining in the bloodstream is modeled by $M(h)=250(0.88)^h$ , where $M$ is in milligrams and $h$ is time in hours. In $M(h)=250(0.88)^h$ , what does the 0.88 represent?

The medication increases by 88% each hour.
The medication decreases by 0.88 milligrams each hour.
Each hour, 88% of the medication from the previous hour remains (a 12% decrease per hour). (correct answer)
The initial dose is 0.88 milligrams.

Explanation: This question tests your ability to interpret the parameters in linear and exponential functions and understand what they mean in real-world contexts. In exponential functions y = a·b^x, the parameter a is the initial value (what y equals when x = 0, because b⁰ = 1), representing the starting amount. The base b is the growth factor (if b > 1) or decay factor (if 0 < b < 1)—it's what you multiply by each time x increases by 1. To find the percent rate: r = b - 1 (giving positive for growth, negative for decay). For P = 500(1.08)^t, a = 500 is initial population, b = 1.08 means multiply by 1.08 yearly (8% growth), so r = 0.08 = 8% annual increase. In this medication model M(h)=250(0.88)^h, the parameter 0.88 is the decay factor b, meaning each hour 88% of the previous amount remains, which is a 12% decrease per hour, with no units as it's a multiplier but contextualized to hourly decay in milligrams. Choice B correctly interprets the parameter 0.88 as 88% of the medication remaining each hour (a 12% decrease per hour). A common mistake, like in choice A, is treating the decay as a linear subtraction—exponentials multiply, so interpret b as the retention factor and calculate percent change as (1-b)*100% for decay. Linear parameter interpretation checklist: (1) identify m (slope) and b (y-intercept) from y = mx + b form, (2) determine units: slope has ratio units (output per input), intercept has output units, (3) interpret m as 'rate of change' or 'amount per unit,' (4) interpret b as 'initial value when x = 0' or 'fixed amount.' Example: y = 15x + 50 for cost vs items → m = 15 $/item (price per item), b = 50$ (starting fee). Always state units—they complete the interpretation! Exponential parameter extraction: (1) identify a and base b from y = a·b^x, (2) interpret a as initial value with output units, (3) classify: b > 1 is growth, 0 < b < 1 is decay, (4) calculate percent rate: r = b - 1 (for growth) or r = 1 - b (for decay, stated as positive percent), multiply by 100 for percent. Example: y = 1000(0.95)^t → a = 1000 initial, b = 0.95 < 1 is decay, r = 1 - 0.95 = 0.05 = 5% decay per period. The base tells you the story—learn to read it!

Question 23

Solve the system using the inverse of the coefficient matrix.

[ \begin{cases} 2x+y=5\ 3x+4y=7 \end{cases} ]

Let $A=\begin{bmatrix}2&1\\3&4\end{bmatrix}$ , $X=\begin{bmatrix}x\\y\end{bmatrix}$ , and $B=\begin{bmatrix}5\\7\end{bmatrix}$ . Find $A^{-1}$ and then compute $X=A^{-1}B$ .

$A^{-1}=\frac{1}{5}\begin{bmatrix}4&-1\\-3&2\end{bmatrix},\; X=\begin{bmatrix}\frac{13}{5}\\-\frac{1}{5}\end{bmatrix}$ (correct answer)
$A^{-1}=\frac{1}{5}\begin{bmatrix}4&1\\3&2\end{bmatrix},\; X=\begin{bmatrix}\frac{13}{5}\\-\frac{1}{5}\end{bmatrix}$
$A^{-1}=\frac{1}{11}\begin{bmatrix}4&-1\\-3&2\end{bmatrix},\; X=\begin{bmatrix}\frac{13}{5}\\-\frac{1}{5}\end{bmatrix}$
$A^{-1}=\frac{1}{5}\begin{bmatrix}4&-1\-3&2\end{bmatrix},\; X=\begin{bmatrix}\frac{12}{5}\\frac{1}{5}\end{bmatrix}$

Explanation: This question tests your understanding of using matrix inverses to solve linear systems—a powerful method that reduces solving a system to a single matrix multiplication once the inverse is found. When a system is written as matrix equation AX = B, we solve for X by computing X = A⁻¹B, where A⁻¹ is the inverse of the coefficient matrix. For the 2×2 matrix A = [[2,1],[3,4]], we first find the determinant: det(A) = 2(4) - 1(3) = 8 - 3 = 5. Using the 2×2 inverse formula, A⁻¹ = (1/5)[[4,-1],[-3,2]] (swap diagonal, negate off-diagonal, divide by determinant). Now compute X = A⁻¹B = (1/5)[[4,-1],[-3,2]][[5],[7]] = (1/5)[[4(5)+(-1)(7)],[(-3)(5)+2(7)]] = (1/5)[[20-7],[-15+14]] = (1/5)[[13],[-1]] = [[13/5],[-1/5]]. Choice A correctly shows both the inverse matrix and the solution vector. Choice B has the wrong inverse (didn't negate the off-diagonal elements), Choice C has the wrong determinant in the scalar factor, and Choice D has arithmetic errors in the final multiplication—each step must be precise!

Question 24

For the logarithmic function $q(x)=2\ln(x)+3$ , identify the vertical asymptote and state the domain. (Connect to the parent function $y=\ln x$ .)

Vertical asymptote $y=3$ ; domain $x\in\mathbb{R}$
Vertical asymptote $x=0$ ; domain $x>0$ (correct answer)
Vertical asymptote $x=3$ ; domain $x>3$
Vertical asymptote $x=0$ ; domain $x\ge 0$

Explanation: This question tests your ability to graph logarithmic functions by identifying their vertical asymptotes and domains. Logarithmic functions f(x) = log_b(x) are the inverses of exponentials, so their graphs are reflections across y = x: they have an x-intercept at (1, 0) because log_b(1) = 0, no y-intercept because log_b(0) is undefined, and a vertical asymptote at x = 0 (the y-axis) that the graph approaches as x → 0⁺. The domain is restricted to x > 0 (can't take log of negative or zero), and end behavior is: as x → 0⁺, f(x) → -∞ (graph goes down along the asymptote), and as x → ∞, f(x) → ∞ (graph rises slowly, flattening as it goes). For q(x) = 2ln(x) + 3, the natural logarithm ln(x) has the same properties as any logarithmic function. The vertical asymptote remains at x = 0 (where ln(x) is undefined), and the domain is still x > 0. The coefficient 2 provides vertical stretch, and the +3 shifts the graph up, but neither transformation affects the asymptote or domain. Choice B correctly identifies the vertical asymptote x = 0 and domain x > 0. Choice C incorrectly shifts the asymptote, while Choice A calls it horizontal instead of vertical. Logarithmic domain strategy: The domain of log_b(x - h) is x > h (shift right by h), but for log_b(x) it's always x > 0. Vertical transformations (like 2ln(x) + 3) don't change the domain or vertical asymptote location—they only stretch and shift the output values!

Question 25

Use polynomial division to write $\frac{x^4-3x^2+5}{x^2+1}$ in the form $q(x)+\frac{r(x)}{x^2+1}$ , where $\deg(r)<\deg(x^2+1)$ .

$x^2-4+\dfrac{9}{x^2+1}$ (correct answer)
$x^2-4+\dfrac{9x}{x^2+1}$
$x^2-3+\dfrac{8}{x^2+1}$
$x^2-4+\dfrac{9}{x+1}$

Explanation: This question tests your understanding of polynomial division—rewriting a rational expression as a polynomial quotient plus a proper fraction, just like rewriting $17/5$ as $3 + 2/5$ . The division algorithm for polynomials says any rational expression $a(x)/b(x)$ can be written as $q(x) + r(x)/b(x)$ , where $q(x)$ is the quotient (polynomial part) and $r(x)$ is the remainder with degree strictly less than the divisor's degree. To divide $x^4 - 3x^2 + 5$ by $x^2 + 1$ , divide $x^4$ by $x^2$ to get $x^2$ , multiply by $(x^2 + 1)$ to get $x^4 + x^2$ , subtract to get $-4x^2 + 5$ , then divide $-4x^2$ by $x^2$ to get $-4$ , multiply to get $-4x^2 - 4$ , and subtract to yield $9$ as remainder. Choice A correctly divides to get quotient $x^2 - 4$ and remainder $9$ with $\deg(r) < \deg(x^2 + 1)$ . Choice C likely results from a miscalculation in the subtraction, getting $-3$ instead of $-4$ and $8$ instead of $9$ . Before dividing, always check: can you factor the numerator and cancel with the denominator? The degree requirement ( $\deg$ of remainder less than $\deg$ of divisor) tells you when to stop dividing: if dividing by $(x^2 + 1)$ (degree 2), remainder can be degree 1 or 0.

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