Polynomial Graph Behavior - Algebra 3

1. Let $g(x)=-x(x-4)^3(x+1)^4$. Which statement correctly describes the graph?
   Answer: The graph falls to the left and right, touches at $x=-1$, crosses at $x=0$ and $x=4$, and is positive on $(0,4)$. Whenever you see a polynomial already factored, start by finding the degree, the leading coefficient, and the multiplicity of each zero. Here $g(x)=-x(x-4)^3(x+1)^4$ has degree $1+3+4=8$ and leading coefficient $-1$, so the graph falls to the left and falls to the right. Next, look at each zero: $x=0$ has multiplicity $1$ (odd), so it crosses; $x=4$ has multiplicity $3$ (odd), so it also crosses; $x=-1$ has multiplicity $4$ (even), so it only touches. That already identifies the correct description: falls both ways, touches at $-1$, crosses at $0$ and $4$. To confirm sign on $(0,4)$, test $x=1$: $-1(1-4)^3(2)^4$ is positive—so yes. In fact, $(0,4)$ is the only positive interval. The distractor saying "rises to the left" has end behavior backwards, treats $x=4$ as a touch instead of a cross, and gives the wrong sign interval. The distractor saying "crosses at $x=-1$" and "touches at $0$ and $4$" swaps the odd/even multiplicities, making $x=4$ touch and sign negative. The distractor saying "touches at $-1$ and $4$" again mistakes the odd multiplicity at $4$, and it also claims the graph is negative on $(0,4)$, which fails the test value. Study takeaway: for each zero, ask "odd multiplicity?" If yes, cross;if even, touch. Then determine end behavior from degree and leading coefficient. For intervals, plug in one test point per region—don't rely on memory.
2. A polynomial function $f$ has x-intercepts at exactly $x=-3$, $x=1$, and $x=4$. Its graph crosses the x-axis at $x=-3$ and $x=4$, but touches, rather than crosses, the x-axis at $x=1$. Also, $f(x)\to -\infty$ as $x\to -\infty$ and $f(x)\to -\infty$ as $x\to \infty$. Which of the following could be $f$ if $f(0)=-24$?
   Answer: No such polynomial function exists. Because the zero at $4$ is a crossing and $f(x)\to -\infty$ as $x\to\infty$, the graph is negative just to the right of $4$ and positive just to the left of $4$. With no zeros between $1$ and $4$, $f$ is positive on $(1,4)$. Since the zero at $1$ is a touch, the sign does not change there, so $f$ is also positive on $(0,1)$. Therefore $f(0)>0$, so no polynomial satisfying the stated zero and end-behavior conditions can have $f(0)=-24$. Among the listed forms, the coefficient $2$ gives $f(0)=-24$ but has a positive leading coefficient (wrong end behavior), while the $-2$ and $-\frac12$ forms give $f(0)=24$ and $6$.
3. Let $p(x)=(2x-1)^3(3x-x^2)^2(x^3+2)$. Which statement about the graph of $p$ is true?
   Answer: The degree is $10$, and $p(x)\to\infty$ as $x\to-\infty$ and as $x\to\infty$. Whenever you see a polynomial in factored form, find the degree by adding the degrees of each factor, and determine end behavior using the degree's parity and the sign of the leading coefficient. Here, $(2x-1)^3$ contributes degree $3$, $(3x-x^2)^2$ contributes degree $4$ because the inside has degree $2$ and is squared, and $(x^3+2)$ contributes degree $3$. So the total degree is $3+4+3=10$. The leading coefficient is positive: from $(2x-1)^3$ it's $2^3=8$, from $(3x-x^2)^2$ the leading term is $x^4$, and from $(x^3+2)$ it's $x^3$. Since the degree is even and the leading coefficient is positive, $p(x)\to\infty$ as $x\to-\infty$ and as $x\to\infty$. That matches the statement: degree $10$, both ends going to $+\infty$. The choices saying degree $8$ come from miscounting — perhaps treating $(3x-x^2)^2$ as degree $2$ or forgetting the $(x^3+2)$ factor entirely. The choice saying degree $10$ but $p(x)\to-\infty$ as $x\to\infty$ confuses the sign: the squared factor is always nonnegative, and its leading term is positive, so the overall leading coefficient is not negative. End behavior is controlled only by the highest-degree term, not by the lower-degree details. Study tip: for end behavior, always compute leading coefficient and degree from the factored form first — then apply: even degree with positive leading coefficient means both ends rise.
4. A 6th-degree polynomial $p$ has a negative leading coefficient and real zeros at $-2,\ 0,\ 3,$ and $5$. Which statement must be true?
   Answer: $p(x)\to-\infty$ as $x\to-\infty$ and as $x\to\infty$. When you see a polynomial with a specified degree and leading coefficient, first isolate what those two facts alone determine: end behavior. A 6th-degree polynomial is even-degree, so both ends go in the same direction. With a negative leading coefficient, both ends must point downward: $p(x)\to -\infty$ as $x\to\pm\infty$. That is the statement that must be true. The given zeros matter for $x$-intercepts, but they do not control the ends. The "exactly five turning points" choice is a trap: a degree 6 polynomial can have at most $6-1=5$ turning points, but multiplicities or non-real zeros can make fewer. Similarly, "crosses the $x$-axis at all four given zeros" is not guaranteed, because a zero with even multiplicity makes the graph touch the axis and bounce, not cross. Finally, $p(x)<0$ on $(0,3)$ assumes a particular sign pattern; but if the zeros at 0 or 3 have even multiplicity, the sign on that interval can flip. For example, $p(x)=-(x+2)x^2(x-3)^2(x-5)$ is degree 6 with the stated zeros and a negative leading coefficient, yet it is positive on $(0,3)$. Study tip: for end-behavior questions, ask only two things — is the degree even or odd, and is the leading coefficient positive or negative? Those determine the ends; zeros and multiplicities shape the middle.
5. Let $f(x)=(2x^2+3)(4-x)^3(x+1)$. Which statement about the graph is true?
   Answer: As $x\to -\infty$, $f(x)\to -\infty$; as $x\to \infty$, $f(x)\to -\infty$; the graph crosses at $x=-1$ and $x=4$. The degree is $2+3+1=6$ and the leading coefficient is $(2)(-1)(1)=-2$, so both ends fall. The factor $2x^2+3$ has no real zeros, $(x+1)$ gives a crossing at $-1$, and $(4-x)^3$ gives a crossing at $4$ because the multiplicity 3 is odd. Statements with opposite end behavior or a touch at a zero of odd multiplicity are incorrect.
6. Let $f(x)=a(x+2)(x-1)^2(x-4)^2$. If the graph has $y$-intercept $16$, which statement is true?
   Answer: The graph falls to the left and rises to the right; it crosses at $x=-2$ and touches at $x=1$ and $x=4$. The $y$-intercept is $f(0)=a(2)(1)(16)=32a$. Setting $32a=16$ gives $a=\frac12>0$. The polynomial has degree $1+2+2=5$ and positive leading coefficient, so it falls to the left and rises to the right. The zero at $-2$ is simple and is a crossing; the zeros at $1$ and $4$ have multiplicity 2 and are touches.
7. Let $f(x)=-2(x-1)^3(x+2)^2(x-3)$. Which statement about the graph of $y=f(x)$ is true?
   Answer: As $x\to -\infty$, $f(x)\to -\infty$; as $x\to \infty$, $f(x)\to -\infty$; the graph crosses at $x=1$ and $x=3$, touches at $x=-2$, and has $y$-intercept $-24$. The degree is $3+2+1=6$, even, and the leading coefficient is $-2$, so both ends fall. The factor $(x+2)^2$ gives a bounce at $-2$, while $(x-1)^3$ and $(x-3)$ give crossings at $1$ and $3$. Also, $f(0)=-2(-1)^3(2)^2(-3)=-24$. Statements that claim opposite end behavior, reverse the tangencies, or give a positive $y$-intercept are incorrect.
8. Let $f(x)=-(x+2)(x-1)^2(x-4)^3$. Which statement about the graph of $f$ is true?
   Answer: At $x=1$, the graph touches the $x$-axis and turns upward, so $f$ has a local minimum there. The degree is $1+2+3=6$, and the leading coefficient is $-1$, so the graph falls to $-\infty$ at both ends. The factor $x+2$ gives a crossing at $-2$, changing the sign from negative to positive. The factor $(x-1)^2$ has even multiplicity, so the graph touches but does not cross the axis at $1$; since $f$ is positive just to both sides of $1$ and $f(1)=0$, this point is a local minimum. The statement about $x=4$ is wrong because the graph crosses there and the sign changes from positive to negative, so it is not positive just to the right. The end-behavior statement is wrong because both ends go to $-\infty$, not one to $\infty$. The interval $(1,4)$ is above the axis, not below.
9. Consider $f(x)=(x^2+1)(x+2)^3(3-x)$. A student makes these four claims. Which claim is false?
   Answer: The highest-degree term is $x^3$, so the graph rises to the left and falls to the right. The true degree is $2+3+1=6$, and the leading term is $(x^2)(x^3)(-x)=-x^6$. Thus both ends fall, not rise left and fall right. The crossings at $-2$ and $3$, the positive sign on $(-2,3)$, and the lack of real zeros from $x^2+1$ are all correct.
10. Let $g(x)=a(x-1)(x-2)^2(x-3)^3$. If $g(-1)=-32$, which statement describes the graph of $g$?
    Answer: Both ends fall; the graph crosses at $x=1$ and $x=3$, bounces at $x=2$, and has $y$-intercept $-3$. Computing $g(-1)=a(-2)(-3)^2(-4)^3=1152a=-32$ gives $a=-\frac{1}{36}$. The degree is $1+2+3=6$, even, and the leading coefficient is negative, so both ends fall. The zeros at $1$ and $3$ have odd multiplicities and are crossings, while the zero at $2$ has even multiplicity and is a bounce. The $y$-intercept is $a(-1)(4)(-27)=-3$.
11. Suppose a polynomial $p$ has only the real zeros $-4,\ 2,\ 7$. If $p(x)>0$ on $(-\infty,-4)$, $p(x)<0$ on $(-4,2)$, $p(x)>0$ on $(2,7)$, and $p(x)<0$ on $(7,\infty)$, which conclusion follows?
    Answer: $p$ has odd degree and a negative leading coefficient, and all three zeros have odd multiplicity. Whenever you see a sign chart for a polynomial, translate it into two things: crossings and end behavior. A sign change at a zero means the factor has odd multiplicity; no sign change would mean even multiplicity. Here the signs go $+,-,+,-$ across $-4,2,7$, so every zero is a crossing: all three zeros have odd multiplicity. Next look far left and far right. The chart gives $p(x)>0$ as $x\to-\infty$ and $p(x)<0$ as $x\to\infty$. Since the right tail is negative, the leading coefficient must be negative. If the degree were even, anegative leading coefficient would make both ends negative; but the left tail is positive. If the degree were odd, anegative leading coefficient makes left positive and right negative, exactly what is given. Thus degree odd, leading coefficient negative, all three zeros odd multiplicity— that matches the choice stating odd degree, negative leading coefficient, and all three zeros odd multiplicity. The other choices fail: "odd degree and a positive leading coefficient" would make the right tail positive, not negative, and at $2$ the sign changes from negative to positive, so that zero cannot have even multiplicity."Even degree and a positive leading coefficient" would make both ends positive, contradicted by the negative right tail."Even degree and a negative leading coefficient" would make both ends negative, contradicted by the positive left tail; also the zeros at $-4$ and $7$ change sign, so they must have odd, not even, multiplicity. Remember the pattern: sign change = odd multiplicity; no sign change = even multiplicity. For end behavior, opposite end signs mean odd degree, and the right tail sign tells you the leading coefficient sign.
12. Let $h(x)=(x^2+3)(x-2)^2(x+1)^4$. Which statement about the graph is true?
    Answer: The graph rises to both the left and the right, touches at $x=-1$ and $x=2$, never crosses the $x$-axis, and has $y$-intercept $12$. The degree is $2+2+4=8$, even, and the leading coefficient is positive, so both ends rise. The factor $x^2+3$ has no real zeros. The zeros at $-1$ and $2$ have even multiplicities, so the graph touches the axis at both and never crosses it. The $y$-intercept is $(3)(4)(1)=12$.
13. Which of the following polynomial functions has a graph that rises to the left, falls to the right, touches the x-axis at $x=0$, and crosses the x-axis at $x=1$?
    Answer: $p(x)=-x^2(x^2+1)(x-1)$ The correct polynomial is $p(x)=-x^2(x^2+1)(x-1)$. Its degree is $2+2+1=5$ and its leading coefficient is $-1$, so the graph rises to the left and falls to the right. The factor $x^2$ gives even multiplicity at $0$, so the graph touches the axis there; the factor $x-1$ gives odd multiplicity at $1$, so the graph crosses there. The factor $x^2+1$ has no real zeros and does not add x-intercepts. The option $-x(x^2+1)(x-1)^2$ has the same end behavior but reverses the touch/cross behavior. The option $x(x^2+1)(x-1)^2$ has opposite end behavior. The option with both $x^2$ and $(x-1)^2$ makes both intercepts touches and has even degree, so both ends fall.
14. Consider $f(x)=(x^2+1)(x+1)^2(x-2)^3$. Which statement correctly describes where the graph lies relative to the $x$-axis?
    Answer: It is below the $x$-axis on $(-\infty,-1)\cup(-1,2)$ and above on $(2,\infty)$. The factor $x^2+1$ is always positive, and $(x+1)^2$ is positive away from $-1$. The sign of $f$ is therefore controlled by $(x-2)^3$. For $x<2$, this factor is negative, so $f<0$ on both $(-\infty,-1)$ and $(-1,2)$. For $x>2$, it is positive, so $f>0$. The graph touches the axis at $-1$ and crosses at $2$.
15. The only real zeros of $p$ are $-3,\ 0,$ and $4$, with multiplicities $1,\ 2,$ and $1$, respectively. If $p(1)>0$, which statement about the graph is true?
    Answer: The leading coefficient is negative, so $p(x)\to-\infty$ as $x\to-\infty$ and as $x\to\infty$. Whenever you see zeros, multiplicities, and a test point like $p(1)>0$, your first move is to write the polynomial in factored form: $p(x)=a(x+3)x^2(x-4)$ The multiplicities become the exponents, and the degree is $1+2+1=4$, which is even. Now use the given sign condition: $p(1)=a(4)(1^2)(-3)=-12a>0$ so $a<0$. That means the leading coefficient is negative. Since the degree is even, a negative leading coefficient sends both ends of the graph downward: $p(x)\to-\infty$ as $x\to-\infty$ and as $x\to\infty$. This matches the statement that the leading coefficient is negative and both ends go to $-\infty$. The choices claiming the leading coefficient is positive contradict $p(1)>0$, so they cannot be true. The choices that describe the two ends going in opposite directions — whether positive leading coefficient with left end going down and right end going up, or negative leading coefficient with left end going up and right end going down — are actually describing odd-degree end behavior. This polynomial has degree $4$, so both ends must go the same direction. On this exam, remember the pairing: even degree means same-direction ends, odd degree means opposite ends. Then use a test point to decide whether both ends go up or down.
16. A polynomial's graph falls to the left, rises to the right, crosses the x-axis at $x=-2$, touches the x-axis at $x=5$, and has y-intercept $50$. Which equation could define this polynomial?
    Answer: $p(x)=(x+2)(x-5)^2$ When you see a polynomial's end behavior, crossing, and touching, think about multiplicity: an odd exponent means the graph crosses, an even exponent means it touches. Here, falling left and rising right means an odd degree with a positive leading coefficient. Crossing at $x=-2$ requires a single factor $(x+2)$, not squared. Touching at $x=5$ requires $(x-5)^2$. So the skeleton is $a(x+2)(x-5)^2$, with $a>0$. The y-intercept is $p(0)=a(2)(25)=50a$, and since the intercept is $50$, $a=1$. Thus the polynomial is $(x+2)(x-5)^2$. The choice $(x+2)^2(x-5)$ reverses the multiplicity roles: it would touch at $-2$ and cross at $5$, opposite to the description, and its y-intercept is $-20$. The choice $-(x+2)(x-5)^2$ has the correct crossing and touching pattern, but its negative leading coefficient makes it fall right and rise left, and its y-intercept is $-50$. The choice $2(x+2)(x-5)^2$ also has correct crossing/touching and positive end behavior, but its y-intercept is $2\cdot 50=100$, not $50$. Your takeaway: when graphing polynomials, convert "crosses" to an odd-powered factor and "touches" to an even-powered factor, then use the y-intercept to solve for the leading coefficient. This prevents all three common traps: wrong multiplicity, wrong sign, and wrong stretch factor.
17. Let $g(x)=-\frac12(x-1)(x+3)^2(x^2+4)$. Which statement about the graph is true?
    Answer: The graph rises to the left, falls to the right, touches at $x=-3$, crosses at $x=1$, and has $y$-intercept $18$. The degree is $1+2+2=5$ and the leading coefficient is $-\frac12$, so the graph rises to the left and falls to the right. The factor $x+3$ is squared, giving a touch at $-3$; $x-1$ is linear, giving a crossing at $1$; and $x^2+4$ is always positive. The $y$-intercept is $-\frac12(-1)(9)(4)=18$.
18. Let $f(x)=4x^3-12x^2+9x$. Which statement describes its graph?
    Answer: The graph falls to the left, rises to the right, crosses at $x=0$, and touches at $x=3/2$. Factoring gives $f(x)=x(2x-3)^2$. The degree is 3 and the leading coefficient is positive, so the graph falls to the left and rises to the right. The zero at $x=0$ has multiplicity 1, so the graph crosses; the zero at $x=3/2$ has multiplicity 2, so the graph touches. Descriptions that reverse these multiplicities or choose a different end behavior are incorrect.
19. Let $h(x)=-(x-2)(x+3)^2(x+1)(x-5)^2$. Which statement about the graph of $h$ is true?
    Answer: The graph falls to the left and right; it crosses at $x=-1$ and $x=2$ and touches at $x=-3$ and $x=5$. Whenever you analyze a polynomial graph, let two ideas guide you: the sign of the leading coefficient and degree determine end behavior, and the multiplicity of each factor determines whether the graph crosses or touches at its zero. Here $h(x)$ has total degree $1+2+1+2=6$, and the leading coefficient is negative because of the minus sign in front. An even-degree polynomial with a negative leading coefficient falls on both ends. Now inspect zeros: $x=-1$ comes from $(x+1)$ with multiplicity 1, so the graph crosses; $x=2$ comes from $(x-2)$ with multiplicity 1, so it crosses. But $x=-3$ comes from $(x+3)^2$ and $x=5$ comes from $(x-5)^2$, both multiplicity 2, so the graph touches at those points instead of crossing. Thus the true statement is: the graph falls to the left and right; it crosses at $-1$ and$2$ and touches at $-3$ and$5$. Each wrong choice misses one piece. The choice that says rises left and falls right mistakes the end behavior as if the leading coefficient were negative with odd degree, and it also swaps which zeros cross and touch. The choice that says falls left/right but crosses at $-3,5$ and touches at $-1,2$ gets end behavior right but assigns the wrong behavior to each zero. The choice that says rises to the left and right correctly identifies the crossing/touching zeros but ignores the negative leading coefficient, which forces both ends to fall. By checking degree, leading sign, and multiplicity, you can eliminate all of these traps.
20. Which statement best explains why $f(x)=-x^5-x^3-x-1$ must have at least one real zero?
    Answer: As $x\to -\infty$, $f(x)\to \infty$, and as $x\to \infty$, $f(x)\to -\infty$; since polynomial graphs are continuous, the graph must cross the $x$-axis somewhere. The leading term is $-x^5$, so the graph rises to the left and falls to the right. A continuous graph that goes from positive values far to the left to negative values far to the right must cross the $x$-axis at least once. Turning-point count alone does not force a crossing, beginning below and continuing downward does not force a crossing, and the graph does not cross at $x=0$ because $f(0)=-1$.

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