In the function above, , , and are positive constants, and . Which choice must be true? I. For every real number such that , . II. As , . III. The graph of crosses the -axis at and at .

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Super-Hard SAT Math Question 135 | Free SAT Prep | Aniko

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Question 135·200 Super-Hard SAT Math Questions·Advanced Math

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f(x)=-(x-a)^2(x+b)(x-c)

In the function above, $a$ , $b$ , and $c$ are positive constants, and $a<c$ . Which choice must be true?

I. For every real number $x$ such that $-b<x<c$ , $f(x)>0$ .

II. As $x\to\pm\infty$ , $f(x)\to -\infty$ . ЅAТ prep b‍у Аniкo.ai

III. The graph of $f$ crosses the $x$ -axis at $x=-b$ and at $x=c$ .

For factored polynomials with parameters, (1) locate zeros from factors and use multiplicity (even = touch/turn, odd = cross), (2) test “for every $x$ in an interval” claims by checking whether any root lies inside the interval, and (3) get end behavior from degree and leading coefficient (here, the dominant term is $-x^4$ ). © Аniko

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Use the fact that $a$ is between $-b$ and $c$

Because $b>0$ and $a,c>0$ with $a<c$ , the number $a$ lies in the interval $(-b,c)$ . Try evaluating $f(x)$ at $x=a$ when checking statement I.

End behavior comes from the leading term

Look at the degree and leading coefficient: $(x-a)^2(x+b)(x-c)$ behaves like $x^4$ for large $|x|$ , and there is a negative sign in front. From аniкo.ai

Crossing vs touching depends on multiplicity

A linear factor to the first power (odd multiplicity) means the graph crosses at that intercept; a squared factor (even multiplicity) means it touches/turns.

Desmos Guide

Enter the function with sliders

In Desmos, enter:

f(x)=-(x-a)^2*(x+b)*(x-c)

Create sliders for $a$ , $b$ , and $c$ .

Choose values satisfying the constraints

Set $a>0$ , $b>0$ , and $c>a$ (for example, $a=2$ , $b=3$ , $c=5$ ). ЅAT рr‌ер‌ ‌bу ⁠Аn‌ikо.‍ai

Check statement I visually

Zoom to show the interval from $x=-b$ to $x=c$ . Notice that the graph hits the $x$ -axis at $x=a$ (which lies between $-b$ and $c$ ), so $f(x)$ is not strictly above $0$ throughout that open interval.

Check statements II and III visually

Zoom out to observe both ends of the graph as $x$ becomes very large positive and very large negative. Then zoom near $x=-b$ and $x=c$ and observe whether the curve crosses the $x$ -axis at those intercepts.

Step-by-step Explanation

Record key structural facts from the factorization

From $f(x)=-(x-a)^2(x+b)(x-c)$ :

Zeros occur at $x=a$ , $x=-b$ , and $x=c$ .
The factor $(x-a)^2$ gives an even multiplicity at $x=a$ .
The factors $(x+b)$ and $(x-c)$ each have power $1$ , so they give odd multiplicity at $x=-b$ and $x=c$ .

Also, since $a>0$ , $c>a$ , and $b>0$ , we have $-b<0<a<c$ , so $a$ is inside the interval $(-b,c)$ .

Test statement I by checking what happens at $x=a$

Statement I claims that for every $x$ with $-b<x<c$ , we have $f(x)>0$ .

But $a$ satisfies $-b<a<c$ , and Wri‌t⁠tе‌n‍ b‍у Aniко

f(a)=-(a-a)^2(a+b)(a-c)=-(0)^2(a+b)(a-c)=0.

So there is at least one $x$ in $(-b,c)$ where $f(x)$ is not greater than $0$ , which is enough to rule out statement I.

Determine end behavior for statement II

For large $|x|$ , the dominant term comes from multiplying the leading parts of each factor:

(x-a)^2(x+b)(x-c) \sim x^2\cdot x\cdot x = x^4.

Because of the negative sign in front, $f(x)\sim -x^4$ . Since $-x^4\to -\infty$ as $x\to\infty$ and also as $x\to -\infty$ , statement II holds.

Use multiplicity to analyze crossing for statement III

At $x=-b$ , the factor $(x+b)$ has odd multiplicity (power $1$ ), so the graph crosses the $x$ -axis there.

At $x=c$ , the factor $(x-c)$ also has odd multiplicity (power $1$ ), so the graph crosses the $x$ -axis there.

(Meanwhile, $(x-a)^2$ would make the graph touch/turn at $x=a$ , not cross.) Thus statement III holds.

Select the choice that matches which statements must be true

Statement I is false, while statements II and III are true. Therefore, the correct choice is II and III only.

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