Question 129·200 Super-Hard SAT Math Questions·Advanced Math

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Function $f$ is a quadratic function. The graph of $y=f(x)$ in the $xy$ -plane has a vertex at $(-2,7)$ and passes through the point $(1,-5)$ . The graph of $y=f(x)$ has a $y$ -intercept at $(0,a)$ .

A new function $p$ is defined by

$p(x)=\frac{1}{2}f(3x+6)-4$ .

The graph of $y=p(x)$ has a $y$ -intercept at $(0,b)$ . Which choice is the positive difference between $a$ and $b$ ?

A vertex is a quadratic graph’s turning point. When a vertex and one other point are given, use vertex form and fit its one unknown scale in Desmos. For a transformed function’s $y$ -intercept, put $0$ into the new rule, evaluate the whole input inside $f$ , and then find the positive difference. The new intercept alone isn’t that difference.

Hints

Hint 1
A quadratic with vertex $(h,k)$ has vertex form $f(x)=c(x-h)^2+k$ . The square is $0$ at $x=h$ , so the given vertex fixes everything except $c$ . What form does $(-2,7)$ give you?
Hint 2
A point $(u,v)$ on a graph means $f(u)=v$ . Use the other given point to find $c$ : substituting the vertex into vertex form makes the squared part $0$ , so it can’t tell you the scale.
Hint 3
At a $y$ -intercept, the input is $0$ . Put $x=0$ into $p(x)$ first. What whole input inside $f$ do you get? Apply the factor and subtraction outside $f$ afterward.

Step-by-step

Build the quadratic, then evaluate both intercepts
Step 1
Write the quadratic in vertex form
The vertex $(-2,7)$ is where the quadratic turns. Vertex form puts a squared term at $0$ when $x=-2$ , so write $f(x)=c(x+2)^2+7$ . The scale $c$ is the only number still unknown.
Step 2
Turn the other point into an equation
Passing through $(1,-5)$ means putting in $1$ gives an output of $-5$ . Substitute that point into the form you wrote: $-5=c(1+2)^2+7$ . Unlike the vertex, this point can determine $c$ because its squared term isn’t $0$ .
Step 3
Find the scale in Desmos
Type $c(1+2)^2+7\sim-5$ . The $\sim$ tells Desmos to fit $c$ so the point works. Under PARAMETERS, Desmos shows $c\approx-1.3333$ . Type $c$ on the next line and tap its fraction button to read the exact value $c=-\frac43$ .
Step 4
Find the original y-intercept
A $y$ -intercept has input $0$ , so $a=f(0)$ . Type the function using Desmos’s fitted $c$ , then type $f(0)$ . Desmos gives $a=\frac53$ ; use the fraction button if it first shows a decimal.
Step 5
Find the new y-intercept
The new graph also has input $0$ at its $y$ -intercept. Substituting into its rule gives $p(0)=\frac12f(3(0)+6)-4$ . Simplify the input inside $f$ : $p(0)=\frac12f(6)-4$ . The new intercept uses $f(6)$ , not $f(0)$ . Type the given rule for $p$ , then $p(0)$ . Desmos gives $b=-\frac{259}{6}$ .
Step 6
Take the positive difference
The positive difference is the distance between the two intercept values, so type $|f(0)-p(0)|$ . Desmos shows $\frac{269}{6}$ (tap the fraction button if needed). The $y$ -intercepts differ by $\frac{269}{6}$ . Choice A.

Question 129·200 Super-Hard SAT Math Questions·Advanced Math

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