is a factor of the quadratic function . Let . If the minimum of is and is a positive zero of , what is the value of ?

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Super-Hard SAT Math Question 136 | Free SAT Prep | Aniko

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Question 136·200 Super-Hard SAT Math Questions·Advanced Math

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$(2x+3)$ is a factor of the quadratic function $y=f(x)$ . Let $g(x)=f(4x-1)$ . If the minimum of $y=g(x)$ is $\left(\frac{19}{32}, k\right)$ and $x=c$ is a positive zero of $f(x)$ , what is the value of $c$ ?

When a quadratic’s factor gives one zero, immediately solve for that root. If vertex information is given through a transformed function (like $g(x)=f(4x-1)$ ), first translate the vertex $x$ -coordinate back to the original function by undoing the input transformation. Then use symmetry: the vertex $x$ -coordinate is the midpoint of the two zeros, so set $\frac{r+c}{2}=h$ and solve for $c$ efficiently.

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Turn the factor into a zero

If $(2x+3)$ is a factor of $f(x)$ , then $f(x)=0$ when $2x+3=0$ . Solve for that $x$ -value.

Translate the vertex information from $g$ to $f$

Since $g(x)=f(4x-1)$ , the $x$ -value in $g$ corresponds to the input $4x-1$ in $f$ . Use the given minimum point of $g$ to find the $x$ -coordinate of the vertex of $f$ .

Use symmetry (midpoint of zeros)

For a quadratic, the vertex $x$ -coordinate is the average of the two zeros: $\frac{r+c}{2}=h$ . Plug in the known zero $r$ and the vertex $x$ -coordinate $h$ , then solve for $c$ .

Desmos Guide

Enter the known root and the given vertex x-value for $g$

In Desmos, enter

r=-3/2
xg=19/32

Convert the vertex x-value from $g$ to the vertex x-value of $f$

Enter h=4xg-1. This computes the vertex $x$ -coordinate of $f$ from the vertex $x$ -coordinate of $g$ .

Compute the other zero using the midpoint relationship

Enter c=2h-r. Then select the answer choice that matches the displayed value of c.

Step-by-step Explanation

Find the known zero of $f(x)$ from the factor

Since $(2x+3)$ is a factor of $f(x)$ , one zero occurs when

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

So one zero of $f(x)$ is $r=-\frac{3}{2}$ .

Use the vertex of $g$ to find the vertex $x$ -coordinate of $f$

Because $g(x)=f(4x-1)$ , the input to $f$ is $4x-1$ .

If $g$ has its minimum at $x=\frac{19}{32}$ , then $f$ has its minimum at the input value

h=4\left(\frac{19}{32}\right)-1=\frac{19}{8}-1=\frac{11}{8}.

So the vertex $x$ -coordinate of $f(x)$ is $h=\frac{11}{8}$ .

Set up the midpoint equation for the zeros

For a quadratic with real zeros $r$ and $c$ , the vertex $x$ -coordinate is the average:

\frac{r+c}{2}=h.

Substitute $r=-\frac{3}{2}$ and $h=\frac{11}{8}$ :

\frac{-\frac{3}{2}+c}{2}=\frac{11}{8}.

Solve for $c$

Multiply both sides by 2:

-\frac{3}{2}+c=\frac{11}{4}.

Add $\frac{3}{2}$ to both sides:

\begin{aligned} c &= \frac{11}{4}+\frac{3}{2} \\ &= \frac{11}{4}+\frac{6}{4} \\ &= \frac{17}{4}. \end{aligned}

Therefore, $c=\frac{17}{4}$ .

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