Super-Hard SAT Advanced Math Question 13 | Free SAT Prep

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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 $a+b$ ?

When a factored function is evaluated at another expression, preserve the factor structure instead of expanding the original function. Substitute the new input into each factor, cancel common factors carefully, and then solve the resulting linear factors. Also check whether any original root cannot be reached by the transformed input; that root will not create an x-intercept.

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Use the factored form

Substitute $\frac{2x-1}{x+3}$ for the input in each of the three factors of $f$ .

Look for cancellation

Each substituted factor has denominator $x+3$ . Consider how the outside factor $(x+3)^3$ affects those denominators.

Identify the zero-producing factors

After simplifying, ignore any nonzero constant factor and set each remaining linear factor equal to zero.

Desmos Guide

Use the simplified equivalent expression

Algebra is fastest here. For a Desmos verification, enter $y=(-7)(3x+2)(-2x-13)$ . This expression has the same x-intercepts as $g$ ; the excluded value $x=-3$ is not an x-intercept.

Read the intercepts

Click the two x-intercepts of the graph to obtain their x-coordinates.

Add the coordinates

Enter the sum of the two displayed x-coordinates in a new expression line to verify the requested value.

Step-by-step Explanation

Substitute into the factors of $f$

Replace the input of each factor of $f$ with $\frac{2x-1}{x+3}$ :

\begin{aligned} g(x)&=(x+3)^3\left(\frac{2x-1}{x+3}-2\right) \left(\frac{2x-1}{x+3}+1\right) \left(\frac{2x-1}{x+3}-4\right). \end{aligned}

The three denominators cancel with $(x+3)^3$ .

Simplify the resulting factors

After combining the terms in each numerator,

\begin{aligned} g(x)&=[(2x-1)-2(x+3)]\,[(2x-1)+(x+3)]\\ &\quad\cdot[(2x-1)-4(x+3)]\\ &=(-7)(3x+2)(-2x-13). \end{aligned}

The constant factor $-7$ is never zero. It results because the input $\frac{2x-1}{x+3}$ can never equal $2$ .

Find and add the x-intercepts

Set each variable factor equal to zero:

3x+2=0 \quad\Rightarrow\quad x=-\frac{2}{3}

-2x-13=0 \quad\Rightarrow\quad x=-\frac{13}{2}

Therefore,

a+b=-\frac{2}{3}-\frac{13}{2}=-\frac{43}{6}.

So choice B is correct.

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Step-by-step Explanation

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