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 $x$ increases without bound or decreases without bound, $f(x)$ decreases without bound.

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

A factored polynomial shows where its output is zero and how its graph behaves near those zeros. For a “must be true” question, test any every-input claim at a value that makes a factor zero. Then use the highest-power term for the far ends and check whether each zero makes the graph cross or touch the axis. A squared factor can interrupt an otherwise positive interval.

Hints

Hint 1
The positive constants put $-b$ to the left of $0$ and put $a$ between $0$ and $c$ . Where does that place $a$ relative to the interval in statement I?
Hint 2
A zero is an input where the function’s output is $0$ . At the input you located, what happens to the squared factor $(x-a)^2$ ? Can an output of $0$ satisfy $f(x)>0$ ?
Hint 3
The leading term is the highest-power part of a polynomial; it controls both far ends of its graph. Multiply the $x$ parts of the factors, keeping the outside minus sign. Which way will both ends go?

Step-by-step

Read the factored form
Step 1
Find an input inside the interval
No Desmos needed. The factored form gives exact facts for every allowed choice of constants; one graph would show only chosen values. Since $b>0$ and $0<a<c$ , the inputs line up as:
$-b<0<a<c.$
So $x=a$ lies inside the interval $-b<x<c$ in statement I.
Step 2
Test the every-input claim
At $x=a$ , the squared factor is $0$ :
$f(a)=-(a-a)^2(a+b)(a-c)=0.$
I requires $f(x)>0$ at every input in that interval, so I fails. One zero inside an interval defeats a claim of positive output everywhere.
Step 3
Check both far ends
The leading term, the highest-power term, controls the graph far to either side. Multiply the highest-power parts of the given factors, keeping the outside minus sign:
$-(x^2)(x)(x)=-x^4.$
As $x$ grows without bound in either direction, $x^4$ grows without bound, so $-x^4$ decreases without bound. II is true.
Step 4
Locate the other zeros
The factors $x+b$ and $x-c$ become $0$ at $x=-b$ and $x=c$ , so:
$f(-b)=0,\qquad f(c)=0.$
These are $x$ -intercepts, points where the graph meets the $x$ -axis. Meeting the axis doesn't yet prove crossing: a graph can touch it at a squared factor.
Step 5
Check the factors at those intercepts
Use $-b<0<a<c$ to check the other factors:
$-b-a<0,\quad -b-c<0;\qquad c-a>0,\quad c+b>0.$
So at $-b$ , only $x+b$ is zero; at $c$ , only $x-c$ is zero.
Step 6
Decide whether the graph crosses
Each of those linear factors, factors with $x$ to the first power, changes sign as $x$ passes its zero. The other factors stay nonzero nearby and keep their signs, so $f$ changes sign at both intercepts. The graph crosses at $-b$ and $c$ , making III true. Only II and III must be true. Choice A.

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

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