Question 13·25 Super-Hard SAT Advanced Math Questions·Advanced Math

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Let $f(x)=(x-2)(x+1)(x-4)$ . For $x\ne-3$ , function $g$ is defined by

g(x)=(x+3)^3f\left(\frac{2x-1}{x+3}\right).

The graph of $y=g(x)$ has x-intercepts at $(a,0)$ and $(b,0)$ , where $a$ and $b$ are distinct constants. Which choice gives $a+b$ ?

In a composite function, one expression supplies the input to another function. To find an $x$ -intercept, look for inputs that make the whole function equal $0$ . Graph the function in Desmos to count its intercepts, then use the factored outer function to find their exact coordinates. The outer function’s zeros are possible inner outputs, not automatically $x$ -intercepts.

Hints

Hint 1
An $x$ -intercept is a point where the output is $0$ . Since $x\ne-3$ , the factor $(x+3)^3$ cannot be $0$ . What must be true of the $f$ part of $g(x)$ ?
Hint 2
The zero-product property says a product is $0$ only if at least one factor is $0$ . Apply it to $f$ : which values could the fraction inside $f$ equal? Check whether the fraction can actually reach each value.
Hint 3
The graph gives approximate intercepts. For exact values, set the fraction inside $f$ equal to each reachable zero of $f$ , then solve for $x$ . Add those $x$ -values, not the zeros of $f$ .

Step-by-step

Graph, then find the exact intercepts
Step 1
Count the intercepts
Type both function rules in Desmos, keeping the restriction on $g$ . Click the two points where its graph meets the $x$ -axis. Desmos shows approximately $(-6.5,0)$ and $(-0.667,0)$ . Their $x$ -coordinates make $g(x)=0$ , but the rounded second coordinate isn't exact enough to use in the sum.
Step 2
Find the possible inputs to $f$
Let $u=\frac{2x-1}{x+3}$ , the whole input to $f$ . The given $x\ne-3$ makes $(x+3)^3$ nonzero, so $g(x)=0$ requires $f(u)=0$ . By the zero-product property, one factor of $f(u)=(u-2)(u+1)(u-4)$ must be zero: $u=2,\quad u=-1,\quad\text{or}\quad u=4$ . These are values of $u$ , not yet values of $x$ .
Step 3
Rule out the unreachable zero
Test whether the fraction can equal $2$ : $\frac{2x-1}{x+3}=2$ . Multiply by $x+3$ , which the restriction says isn't zero: $2x-1=2(x+3)$ . Distribute the $2$ : $2x-1=2x+6$ . Subtract $2x$ : $-1=6$ . That's impossible, so $u=2$ gives no intercept. Don't count all three zeros of $f$ as zeros of $g$ .
Step 4
Find the intercept from $u=-1$
Set the fraction equal to $-1$ : $\frac{2x-1}{x+3}=-1$ . Multiply by $x+3$ : $2x-1=-x-3$ . Add $x$ : $3x-1=-3$ . Add $1$ : $3x=-2$ . Divide by $3$ : $x=-\frac{2}{3}$ . This is the exact value behind the graph's rounded $-0.667$ .
Step 5
Find the intercept from $u=4$
Set the fraction equal to $4$ : $\frac{2x-1}{x+3}=4$ . Multiply by $x+3$ : $2x-1=4(x+3)$ . Distribute the $4$ to both terms: $2x-1=4x+12$ . Subtract $2x$ : $-1=2x+12$ . Subtract $12$ : $-13=2x$ . Divide by $2$ : $x=-\frac{13}{2}$ . Neither intercept makes $x+3=0$ , so both are allowed.
Step 6
Add the two intercept coordinates
The question asks for the sum of the $x$ -coordinates. Type $-\frac{2}{3}-\frac{13}{2}$ in Desmos. It displays about $-7.1667$ ; use its fraction button to read $-\frac{43}{6}$ . So $a+b=-\frac{43}{6}$ . Choice B.

Question 13·25 Super-Hard SAT Advanced Math Questions·Advanced Math

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