Question 115·200 Super-Hard SAT Math Questions·Advanced Math

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Function $f$ is defined by

f(x)=(x+13)(x+9)(x+5)(x-2).

Function $g$ is defined by $g(x)=f(x^2-6x)$ . Which choice gives the number of distinct $x$ -intercepts of the graph of $y=g(x)$ ?

A composite function puts one rule’s output into another. Here, the factored outer rule gives possible values of the inner expression, not necessarily values of $x$ . To count $x$ -intercepts, find which of those values the inner quadratic can reach, then count the inputs that produce each one. An upward-opening quadratic reaches its minimum once and a higher value twice.

Hints

Hint 1
An $x$ -intercept has output $0$ . Set the outer function equal to $0$ and use its factors to find the possible inputs to $f$ . Those inputs are values of $x^2-6x$ , not yet values of $x$ .
Hint 2
Complete the square in $x^2-6x$ . A square can’t be negative, so the resulting form shows the smallest value the inner expression can reach. Which possible input to $f$ lies below it?
Hint 3
At a quadratic’s vertex, its turning point, one $x$ produces the minimum. A value above the minimum comes from two $x$ -values, one on each side. Apply that count to each reachable input to $f$ .

Step-by-step

Count inputs through the two functions
Step 1
Turn an intercept into an equation
At an $x$ -intercept, the output is $0$ . Since $g$ puts $x^2-6x$ into $f$ , the zero-output condition is $g(x)=0\quad\Longleftrightarrow\quad f(x^2-6x)=0$ .
Step 2
Find the possible inner outputs
Let $u=x^2-6x$ . A product is $0$ when at least one factor is $0$ , so apply that rule to the given factored form: $(u+13)(u+9)(u+5)(u-2)=0$ , $u=-13,\,-9,\,-5,\text{ or }2$ . These are possible values of $u$ , not four $x$ -intercepts.
Step 3
Find which values the inner expression can reach
Half of $-6$ is $-3$ . Add and subtract $3^2$ to complete the square: $\begin{aligned}u&=x^2-6x\\&=x^2-6x+9-9\\&=(x-3)^2-9\ge -9.\end{aligned}$ A square can’t be negative, so $-13$ is out of reach. Don’t count an $x$ for that outer zero.
Step 4
Count the input at the minimum
For $u=-9$ , put that value into the squared form: $(x-3)^2-9=-9$ . Add $9$ to both sides: $(x-3)^2=0$ . A square is $0$ only when its base is $0$ , so: $x-3=0$ . Add $3$ to both sides: $x=3$ . This branch gives one input, not two.
Step 5
Count the inputs above the minimum
Put each remaining value of $u$ into the squared form, then add $9$ : $\begin{aligned}(x-3)^2-9&=-5\\(x-3)^2&=4\end{aligned}\qquad\begin{aligned}(x-3)^2-9&=2\\(x-3)^2&=11.\end{aligned}$ Each positive square gives two $x$ -values, one on either side of $3$ . The pairs can’t overlap: one $x$ can’t make $u$ equal both $-5$ and $2$ .
Step 6
Check the distinct intercepts in Desmos
Type the given rule as $f(x)=(x+13)(x+9)(x+5)(x-2)$ , then type $g(x)=f(x^2-6x)$ . Click the curve for $g$ , not the curve for $f$ . Its intercepts are at about $x=-0.317,1,3,5,6.317$ . The graph confirms the one-at-the-minimum and two-per-higher-value count: the graph of $y=g(x)$ has 5 distinct $x$ -intercepts. Choice B.

Question 115·200 Super-Hard SAT Math Questions·Advanced Math

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