Super-Hard SAT Advanced Math Question 9 | Free SAT Prep

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Question 9·25 Super-Hard SAT Advanced Math Questions·Advanced Math

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Function $f$ is defined by

f(x)=(x-1)(x-4)(x-9).

Function $g$ is defined by

g(x)=f\left((x-2)^2-1\right).

The graph of $y=g(x)$ has six distinct x-intercepts. Which choice is the product of the x-coordinates of all six x-intercepts?

When a function is composed with a squared expression, first identify the zeros of the outer function. Each outer zero often creates two symmetric x-values. If the question asks for a product or sum of many intercepts, group each symmetric pair and use its sum or product instead of solving for every individual intercept.

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Use the zeros of $f$

For $g(x)$ to equal $0$ , the expression $(x-2)^2-1$ must be equal to one of the three zeros of $f$ .

Keep each pair together

Each zero of $f$ gives two x-values of the form $2+\sqrt{r+1}$ and $2-\sqrt{r+1}$ . Find the product of these two values using a difference of squares.

Combine the results

After finding one product for each of the three zeros of $f$ , multiply those three products to obtain the product of all six x-coordinates.

Desmos Guide

Prefer the algebraic pairing method

Algebra is faster here because the x-intercepts occur in pairs centered at $x=2$ , and the difference-of-squares product gives each pair's product directly.

Graph the composed function for verification

Enter f(t)=(t-1)(t-4)(t-9) and then enter g(x)=f((x-2)^2-1). Graph $y=g(x)$ and click the x-intercepts.

Check the paired structure

The intercepts appear in three pairs, each with an average x-coordinate of $2$ . This confirms the form $2\pm\sqrt{r+1}$ used in the algebraic method.

Step-by-step Explanation

Set the input of $f$ equal to a zero

For $g(x)=0$ , the input to $f$ must equal one of the zeros of $f$ : $1$ , $4$ , or $9$ .

Thus,

(x-2)^2-1=r,

where $r$ is $1$ , $4$ , or $9$ .

Find the product for one pair

For a particular value of $r$ ,

(x-2)^2=r+1.

The two corresponding x-values are $2+\sqrt{r+1}$ and $2-\sqrt{r+1}$ . Their product is

\begin{aligned} \left(2+\sqrt{r+1}\right)\left(2-\sqrt{r+1}\right) &=4-(r+1)\\ &=3-r. \end{aligned}

Multiply the three pair-products

Use $r=1$ , $r=4$ , and $r=9$ :

(3-1)(3-4)(3-9)=2(-1)(-6)=12.

Therefore, the product of all six x-coordinates is $12$ .

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Step-by-step Explanation

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