Algebra 2 Quiz: Using Intersections To Solve Equivalent Functions

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Using Intersections To Solve Equivalent FunctionsQuestion 1 of 20

Solve $2^x=3x+1$ using a table of values (successive approximations). Use the values below to choose the best approximation for the solution, to the nearest tenth.

Table:

At $x=2.0$ : $2^x=4.0$ and $3x+1=7.0$
At $x=2.5$ : $2^x\approx 5.7$ and $3x+1=8.5$
At $x=2.8$ : $2^x\approx 7.0$ and $3x+1=9.4$
At $x=3.0$ : $2^x=8.0$ and $3x+1=10.0$
At $x=3.2$ : $2^x\approx 9.2$ and $3x+1=10.6$
At $x=3.4$ : $2^x\approx 10.6$ and $3x+1=11.2$
At $x=3.5$ : $2^x\approx 11.3$ and $3x+1=11.5$
At $x=3.6$ : $2^x\approx 12.1$ and $3x+1=11.8$

x\approx 3.0

x\approx 3.5

x\approx 2.5

x\approx 11.5

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Algebra 2 Quiz

Algebra 2 Quiz: Using Intersections To Solve Equivalent Functions

Practice Using Intersections To Solve Equivalent Functions in Algebra 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Using Intersections To Solve Equivalent Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Solve $2^x=3x+1$ using a table of values (successive approximations). Use the values below to choose the best approximation for the solution, to the nearest tenth.

Table:

At $x=2.0$ : $2^x=4.0$ and $3x+1=7.0$
At $x=2.5$ : $2^x\approx 5.7$ and $3x+1=8.5$
At $x=2.8$ : $2^x\approx 7.0$ and $3x+1=9.4$
At $x=3.0$ : $2^x=8.0$ and $3x+1=10.0$
At $x=3.2$ : $2^x\approx 9.2$ and $3x+1=10.6$
At $x=3.4$ : $2^x\approx 10.6$ and $3x+1=11.2$
At $x=3.5$ : $2^x\approx 11.3$ and $3x+1=11.5$
At $x=3.6$ : $2^x\approx 12.1$ and $3x+1=11.8$

$x\approx 3.0$
$x\approx 3.5$ (correct answer)
$x\approx 2.5$
$x\approx 11.5$

Explanation: This question tests your understanding that solving the equation f(x) = g(x) is equivalent to finding the x-coordinates where the graphs y = f(x) and y = g(x) intersect—a powerful visual and technological approach to solving equations. The intersection-solution connection works because at an intersection point, both functions have the same y-value: if the graphs meet at (a, b), then f(a) = b and g(a) = b, which means f(a) = g(a)—so x = a solves the equation f(x) = g(x)! Technology makes this powerful for equations that are difficult to solve algebraically, like 2^x = 3x + 1, where we can use tables to approximate solutions by successive narrowing. Using tables, we can approximate by finding where f(x) and g(x) values are closest or where they cross (sign change in f(x) - g(x)), then narrow the interval with smaller steps—for 2^x = 3x + 1, the table shows a crossing between x = 3.5 and x = 3.6. Choice B correctly identifies x ≈ 3.5 as the best approximation to the nearest tenth, since at x = 3.5, the values are very close (11.3 ≈ 11.5), and the actual solution is nearby. Choice A claims x ≈ 3.0, but at x = 3.0, 8.0 < 10.0, and the crossing is later—count all points carefully and narrow intervals to avoid missing the precise spot! Table approximation method: (1) Create tables for both f(x) and g(x) at several x-values, (2) Look for where f(x) approximately equals g(x) or sign changes, (3) Narrow the interval with smaller steps, (4) Continue until desired precision. You're doing great—keep refining those approximations!

Question 2

Use the table to approximate the solution to $f(x)=g(x)$ , where $f(x)=x^3$ and $g(x)=2x+5$ . Choose the best estimate for the solution to the nearest tenth.

Table:

$x$	$f(x)=x^3$	$g(x)=2x+5$
1.5	3.375	8.0
1.8	5.832	8.6
1.9	6.859	8.8
2.0	8.000	9.0
2.1	9.261	9.2
2.2	10.648	9.4

$x\approx 9.2$
$x\approx 1.9$
$x\approx 2.1$ (correct answer)
$x\approx 2.0$

Question 3

Use the table to approximate the solution to $f(x)=g(x)$ , where $f(x)=e^x$ and $g(x)=4x$ . Based on the values below, in which interval does the solution lie?

Values:

At $x=2.1$ : $f(x)\approx 8.17$ , $g(x)=8.40$
At $x=2.2$ : $f(x)\approx 9.03$ , $g(x)=8.80$

Between $2.0$ and $2.1$
There is no solution because $f(x)$ and $g(x)$ are never equal.
Between $2.2$ and $2.3$
Between $2.1$ and $2.2$ (correct answer)

Explanation: This question tests your understanding that solving the equation f(x) = g(x) is equivalent to finding the x-coordinates where the graphs y = f(x) and y = g(x) intersect—a powerful visual and technological approach to solving equations. The intersection-solution connection works because at an intersection point, both functions have the same y-value: if the graphs meet at (a, b), then f(a) = b and g(a) = b, which means f(a) = g(a)—so x = a solves the equation f(x) = g(x)! The table shows that at x = 2.1, f(x) = e^2.1 ≈ 8.17 while g(x) = 4(2.1) = 8.40, so f(x) < g(x). At x = 2.2, f(x) = e^2.2 ≈ 9.03 while g(x) = 4(2.2) = 8.80, so f(x) > g(x). Since f(x) - g(x) changes from negative to positive between x = 2.1 and x = 2.2, the functions must be equal somewhere in this interval—this is the Intermediate Value Theorem in action! Choice A correctly identifies the interval [2.1, 2.2] where the sign change occurs. Choice B suggests [2.0, 2.1], but we'd need table values at x = 2.0 to confirm a sign change there. Choice C suggests [2.2, 2.3], but both functions would need to be evaluated at x = 2.3 to check. Choice D incorrectly claims no solution exists, missing the clear sign change in the table. The table approximation method uses sign changes in f(x) - g(x) to bracket solutions—when the difference switches from negative to positive (or vice versa), a solution lies between those x-values!

Question 4

A table gives values of $f(x)$ and $g(x)$ . Use it to identify the interval that contains a solution to $f(x)=g(x)$ .

Table:

At $x=1.4$ : $f(x)=4.06$ , $g(x)=4.20$
At $x=1.5$ : $f(x)=4.48$ , $g(x)=4.25$
At $x=1.6$ : $f(x)=4.95$ , $g(x)=4.30$

Assume $f$ and $g$ are continuous on this interval. In which interval must a solution occur?

Between $1.4$ and $1.5$ (correct answer)
Between $1.5$ and $1.6$
Between $1.4$ and $1.6$ only (cannot narrow further)
No solution occurs in $[1.4,1.6]$

Question 5

Let $f(x)=\ln(x)$ and $g(x)=x-2$ . Use a graphing calculator (or graphing software) to approximate the solution to $\ln(x)=x-2$ . Give the solution to the nearest hundredth.

$x\approx 1.00$
$x\approx 2.00$
$x\approx 0.16$
$x\approx 3.15$ (correct answer)

Question 6

A student graphs $y = |x - 3| + 1$ and $y = \frac{1}{2}x + 4$ on the same coordinate plane. Based on the intersection points of these graphs, which equation below would have the same solutions?

$|x - 3| + 1 = \frac{1}{2}x + 4$
$|x - 3| = \frac{1}{2}x + 5$
$|x - 3| - \frac{1}{2}x = 3$ (correct answer)
$x - 3 = \frac{1}{2}x + 4$

Explanation: The x-coordinates of intersection points satisfy the equation formed when the two functions are set equal:

|x - 3| + 1 = \frac{1}{2}x + 4

. Subtracting 1 from both sides:

|x - 3| = \frac{1}{2}x + 3

. Subtracting

\frac{1}{2}x

from both sides:

|x - 3| - \frac{1}{2}x = 3

. Choice A is the direct equation but doesn't match any option exactly. Choice B has the wrong constant term. Choice D removes the absolute value bars incorrectly.

Question 7

The rational function $r(x) = \frac{x^2 - 4}{x - 1}$ and the linear function $s(x) = x + 3$ appear to intersect at two points when graphed. However, when solving $\frac{x^2 - 4}{x - 1} = x + 3$ algebraically, why might the number of actual intersection points differ from what the graph suggests?

The rational function has a vertical asymptote that creates the appearance of additional intersection points
Graphing technology cannot accurately display the behavior of rational functions near their asymptotes
The linear function's slope prevents it from intersecting rational functions at more than one point
When cross-multiplying to clear the denominator, extraneous solutions may be introduced that must be checked against the domain restrictions (correct answer)

Explanation: When solving equations involving rational functions, you must be careful about operations that might introduce solutions that don't actually work in the original equation. Let's see what happens when we solve this algebraically. To solve

\frac{x^2 - 4}{x - 1} = x + 3

, you'd cross-multiply to get

x^2 - 4 = (x + 3)(x - 1)

. Expanding the right side gives

x^2 - 4 = x^2 + 2x - 3

, which simplifies to

-4 = 2x - 3

, so

x = -\frac{1}{2}

. However, when you cross-multiply, you're essentially multiplying both sides by

(x - 1)

. If this expression equals zero (when

x = 1

), you've multiplied by zero, which can introduce false solutions that must be checked against the original equation's domain. The correct answer is D. Cross-multiplying can introduce extraneous solutions because you're multiplying by an expression containing the variable. Any solution must be verified in the original equation, and importantly, the rational function

r(x)

is undefined at

x = 1

, so this value is excluded from the domain. Choice A is incorrect because vertical asymptotes don't create the appearance of intersection points—they're places where the function doesn't exist. Choice B is wrong because modern graphing technology accurately displays rational function behavior. Choice C is false because a linear function's slope doesn't inherently limit intersections with rational functions. Remember: whenever you cross-multiply to solve rational equations, always check your solutions in the original equation and verify they don't make any denominator zero.

Question 8

The graphs of $f(x) = \log_2(x + 4)$ and $g(x) = 3 - x$ intersect at point P. Which statement best explains why the x-coordinate of point P is also a solution to $\log_2(x + 4) + x = 3$ ?

At intersection points, both functions have the same slope, so their derivatives are equal
At intersection points, the y-values are equal, so substituting gives $\log_2(x + 4) = 3 - x$ , which rearranges to the given equation (correct answer)
At intersection points, the x-values are equal to the y-values, creating the relationship shown in the equation
At intersection points, the sum of the function values equals the difference, leading to the algebraic form given

Explanation: At intersection points,

f(x) = g(x)

, so

\log_2(x + 4) = 3 - x

. Adding

x

to both sides gives

\log_2(x + 4) + x = 3

. Choice A incorrectly focuses on derivatives. Choice C incorrectly states that x-values equal y-values at intersections. Choice D incorrectly describes the algebraic relationship between the functions.

Question 9

Consider the system where $p(x) = x^3 - 2x^2 + 1$ and $q(x) = 2x - 1$ . A student claims that since both functions pass through the point $(1, 1)$ , the equation $x^3 - 2x^2 + 1 = 2x - 1$ has $x = 1$ as its only solution. What is wrong with this reasoning?

The student made an arithmetic error; the functions don't actually intersect at $(1, 1)$
The student assumed that intersection points are always unique without checking for additional solutions to the equation (correct answer)
The student confused intersection points with critical points of the individual functions
The student incorrectly set up the equation by subtracting instead of adding the functions

Explanation: While both functions do pass through

(1, 1)

, this only confirms that

x = 1

is one solution to

p(x) = q(x)

. The equation

x^3 - 2x^2 + 1 = 2x - 1

simplifies to

x^3 - 2x^2 - 2x + 2 = 0

, which is a cubic equation that could have up to 3 real solutions. The student failed to check for additional intersection points. Choice A is incorrect since

p(1) = q(1) = 1

. Choice C misidentifies the concept. Choice D incorrectly describes the equation setup.

Question 10

A graphing calculator shows that $h(x) = e^{x-2}$ and $k(x) = \frac{6}{x+1}$ intersect at approximately $x = 1.5$ . To verify this intersection point, which calculation should produce a result closest to zero?

$e^{1.5-2} - \frac{6}{1.5+1}$ (correct answer)
$e^{1.5-2} + \frac{6}{1.5+1}$
$\frac{e^{1.5-2}}{\frac{6}{1.5+1}}$
$e^{1.5-2} \times \frac{6}{1.5+1}$

Explanation: At an intersection point, the function values are equal:

h(x) = k(x)

. This means

h(x) - k(x) = 0

. Therefore,

e^{1.5-2} - \frac{6}{1.5+1}

should equal zero if

x = 1.5

is truly an intersection point. Choice B adds the values instead of finding their difference. Choice C finds their ratio, which would equal 1 at intersection. Choice D finds their product, which has no special significance for intersections.

Question 11

The functions $u(x) = 3^x + 1$ and $v(x) = 5 - 2x$ intersect at point Q. If a student uses successive approximations and finds that the x-coordinate of Q is between 1.2 and 1.3, which inequality best represents this approximation process?

$1.2 < x < 1.3$ where $3^x + 1 - (5 - 2x) \approx 0$ (correct answer)
$1.2 < x < 1.3$ where $3^x + 1 + (5 - 2x) \approx 6$
$1.2 < x < 1.3$ where $\frac{3^x + 1}{5 - 2x} \approx 1$
$1.2 < x < 1.3$ where $(3^x + 1)(5 - 2x) \approx 15$

Explanation: The intersection point occurs where

u(x) = v(x)

, or equivalently where

u(x) - v(x) = 0

. Therefore,

3^x + 1 - (5 - 2x) = 0

. The successive approximation process narrows down the interval where this difference is approximately zero. Choice B represents the sum being approximately 6, which isn't the intersection condition. Choice C represents the ratio being approximately 1, which is equivalent but less direct. Choice D represents an arbitrary product relationship.

Question 12

Two functions $f(x) = x^2 - 4x + 5$ and $g(x) = -x^2 + 6x - 1$ are graphed together. At their intersection points, what is the relationship between the solutions of $f(x) = g(x)$ and the solutions of $2x^2 - 10x + 6 = 0$ ?

They are reciprocals of each other due to the algebraic manipulation involved
They are different solutions since the second equation has different coefficients
They are related by a factor of 2, so one set of solutions is twice the other
They are the same solutions since both equations represent the intersection condition (correct answer)

Explanation: When you encounter questions about function intersections, remember that intersection points occur where two functions have equal output values for the same input values. This means you're looking for x-values where

f(x) = g(x)

. To find where these functions intersect, you set them equal:

x^2 - 4x + 5 = -x^2 + 6x - 1

. Moving all terms to one side gives you

x^2 - 4x + 5 + x^2 - 6x + 1 = 0

, which simplifies to

2x^2 - 10x + 6 = 0

. This is exactly the second equation given in the question! The solutions to both equations are identical because they represent the same mathematical condition: finding the x-coordinates where the graphs intersect. Choice A is incorrect because reciprocals aren't involved in this algebraic process—you're simply rearranging terms, not inverting values. Choice B misses the fundamental concept: having different coefficients doesn't mean different solutions when the equations are mathematically equivalent through algebraic manipulation. Choice C incorrectly assumes a proportional relationship between the solution sets, but there's no factor of 2 connecting them—the solutions are exactly the same numbers. The correct answer is D because both equations represent identical intersection conditions, just written in different forms. Study tip: When comparing equations that might represent the same relationships, always check if one can be algebraically transformed into the other. If so, their solutions will be identical, regardless of how different the coefficients initially appear.

Question 13

Solve $2^x=3x+1$ using a table of values (successive approximations). Use the values below to choose the best approximation for the solution, to the nearest tenth.

Table:

At $x=2.0$ : $2^x=4.0$ and $3x+1=7.0$
At $x=2.5$ : $2^x\approx 5.7$ and $3x+1=8.5$
At $x=2.8$ : $2^x\approx 7.0$ and $3x+1=9.4$
At $x=3.0$ : $2^x=8.0$ and $3x+1=10.0$
At $x=3.2$ : $2^x\approx 9.2$ and $3x+1=10.6$
At $x=3.4$ : $2^x\approx 10.6$ and $3x+1=11.2$
At $x=3.5$ : $2^x\approx 11.3$ and $3x+1=11.5$
At $x=3.6$ : $2^x\approx 12.1$ and $3x+1=11.8$

$x\approx 3.5$ (correct answer)
$x\approx 11.5$
$x\approx 3.0$
$x\approx 2.5$

Question 14

Use the table to approximate the solution to $f(x)=g(x)$ , where $f(x)=x^3$ and $g(x)=2x+5$ . Choose the best estimate for the solution to the nearest tenth.

$x\approx 2.0$
$x\approx 2.1$
$x\approx 1.9$
$x\approx 9.2$

Question 15

Define $f(x)=|x-2|$ and $g(x)=x^2-4$ . How many solutions does $f(x)=g(x)$ have? (You may use a graphing calculator to count intersection points.)

0
1
2 (correct answer)
3

Explanation: This question tests your understanding that solving the equation f(x) = g(x) is equivalent to finding the x-coordinates where the graphs y = f(x) and y = g(x) intersect—a powerful visual and technological approach to solving equations. The intersection-solution connection works because at an intersection point, both functions have the same y-value: if the graphs meet at (a, b), then f(a) = b and g(a) = b, which means f(a) = g(a)—so x = a solves the equation f(x) = g(x)! For counting solutions to |x - 2| = x² - 4, we graph the V-shaped absolute value function y = |x - 2| (vertex at (2, 0), going up with slope ±1) and the parabola y = x² - 4 (vertex at (0, -4), opening upward). The absolute value function has two pieces: x - 2 for x ≥ 2 and -(x - 2) = 2 - x for x < 2. For x ≥ 2: x - 2 = x² - 4 gives x² - x - 2 = 0, so (x - 2)(x + 1) = 0, yielding x = 2 (valid since x ≥ 2). For x < 2: 2 - x = x² - 4 gives x² + x - 6 = 0, so (x + 3)(x - 2) = 0, yielding x = -3 (valid since x < 2). Choice C correctly identifies 2 solutions at x = -3 and x = 2. Choice A (0 solutions) would mean the graphs never intersect, Choice B (1 solution) misses one intersection, and Choice D (3 solutions) overcounts—perhaps confusing the number of pieces in the absolute value function with the number of intersections. The graphical approach makes counting solutions visual and intuitive: just count how many times the curves cross!

Question 16

A student graphs $y=f(x)$ and $y=g(x)$ and sees they intersect at the point $(4,7)$ . Which equation must be true?

$f(7)=g(7)$
$f(4)=g(4)$ (correct answer)
$f(4)=g(7)$
$f(x)=g(x)$ for all $x$

Question 17

Let $f(x)=x^2-2$ and $g(x)=2x+1$ . Use the intersection points of the graphs $y=f(x)$ and $y=g(x)$ to solve $f(x)=g(x)$ . Which set of $x$ -values are the solutions?

$x\approx -1.0$ and $x\approx 3.0$ (correct answer)
$x\approx -0.3$ and $x\approx 2.3$
$x\approx -2.3$ and $x\approx 0.3$
$y\approx -0.3$ and $y\approx 2.3$

Explanation: This question tests your understanding that solving the equation f(x) = g(x) is equivalent to finding the x-coordinates where the graphs y = f(x) and y = g(x) intersect—a powerful visual and technological approach to solving equations. The intersection-solution connection works because at an intersection point, both functions have the same y-value: if the graphs meet at (a, b), then f(a) = b and g(a) = b, which means f(a) = g(a)—so x = a solves the equation f(x) = g(x)! This is why we can solve equations graphically: graph both sides as separate functions (y = left side and y = right side), find where they intersect, and read the x-coordinate(s). For solving f(x) = g(x) graphically, we graph both y = f(x) and y = g(x) on the same coordinate system, and the intersection points show where the functions are equal—for example, with f(x) = x² - 2 and g(x) = 2x + 1, they intersect at points with x-coordinates x = -1 and x = 3. Choice B correctly identifies the x-coordinates of intersections as approximately x ≈ -1.0 and x ≈ 3.0, which solve f(x) = g(x) because both functions equal the same y-value at those x-values. Choice D gives the y-coordinates instead of the x-coordinates from the intersection, but the solution to f(x) = g(x) is the x-value where they're equal, not the y-value they both equal—when reading intersections, trace carefully from the crossing point down to the x-axis and read the x-coordinate precisely! The graphical solving recipe: (1) Rewrite equation as f(x) = g(x), (2) Graph y = f(x) and y = g(x) on same axes, (3) Find intersection point(s) visually, (4) Read x-coordinate(s) of each intersection—those are your solutions, (5) Verify by substituting back. Keep practicing graphing to spot all intersections accurately—you've got this!

Question 18

Let $f(x)=x^3$ and $g(x)=4x$ . Use the intersections of $y=f(x)$ and $y=g(x)$ to solve $f(x)=g(x)$ (exact solutions).

$x=-2,0,2$ (correct answer)
$x=-2,2$
$x=0,4$
$x=-4,0,4$

Question 19

The graphs of $y=f(x)$ and $y=g(x)$ intersect at the point $(a,b)$ . Which statement must be true and explains why the solution(s) to $f(x)=g(x)$ come from intersection points?

At an intersection, the $y$ -values are equal, so $f(a)=b$ and $g(a)=b$ , which means $f(a)=g(a)$ . (correct answer)
At an intersection, the $x$ -coordinate $a$ equals the $y$ -coordinate $b$ , so $f(a)=g(a)$ .
At an intersection, the $x$ -values are equal, so $f(x)=g(x)$ for all $x$ .
At an intersection, the slopes are equal, so $f(a)=g(a)$ .

Question 20

Solve $f(x)=g(x)$ approximately by finding where the graphs intersect: $f(x)=\dfrac{x+1}{x-1}$ and $g(x)=2$ . Give the solution(s).

$x=3$ (correct answer)
$x=-1$
$x=1$
$x=2$