Question 15·25 Super-Hard SAT Advanced Math Questions·Advanced Math

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A quadratic function $f$ is defined by $f(x)=ax^2+bx+c$ , where $a$ , $b$ , and $c$ are constants and $a\ne0$ . Function $g$ is defined by $g(x)=f(x+2)-f(x-2)$ . In the $xy$ -plane, the graph of $y=g(x)$ passes through $(1,0)$ . The graph of $y=f(x)$ has two $x$ -intercepts whose $x$ -coordinates have a product of $-5$ , and the $y$ -intercept of $y=f(x)$ is $(0,35)$ . What is the value of $f(3)$ ?

Your answer

A zero value of a shifted difference tells you that two outputs of a quadratic are equal. Equal outputs lie on opposite sides of the axis of symmetry, so they connect the quadratic’s coefficients. Then use the intercept product to find the remaining coefficient and evaluate the function in Desmos. The key trap is treating a zero output of $g$ as a zero output of $f$ .

Hints

Hint 1
The point $(1,0)$ on $y=g(x)$ means $g(1)=0$ . Put $1$ into both inputs inside $f$ . What must be true of the two outputs if their difference is zero?
Hint 2
Equal outputs of a quadratic come from inputs equally far from its axis of symmetry. Find the midpoint of $-1$ and $3$ , then connect it to the axis formula $x=-b/(2a)$ .
Hint 3
The two roots, or inputs where $f(x)=0$ , have a product of $c/a$ , not $c$ . Get $c$ from the $y$ -intercept before using that product to find $a$ .

Step-by-step

Use symmetry and the intercept product
Step 1
Find the constant term
The $y$ -intercept $(0,35)$ gives the output when the input is $0$ . The terms containing $x$ vanish there, so:
$f(0)=c=35.$
Step 2
Turn the point on $g$ into equal outputs
The point $(1,0)$ says $g(1)=0$ . Put $1$ into both shifted inputs:
$f(1+2)-f(1-2)=0.$
Simplify the inputs:
$f(3)-f(-1)=0.$
Their difference is zero, so:
$f(3)=f(-1).$
This does not say $f(3)=0$ .
Step 3
Use symmetry to connect $a$ and $b$
The axis of symmetry is the vertical line halfway between inputs with equal outputs. Equal outputs of a quadratic come from inputs equally far from its axis. Here $-1$ and $3$ are each $2$ units from $1$ , so the axis is $x=1$ . For $ax^2+bx+c$ , the axis is $x=-\frac{b}{2a}$ , giving:
$-\frac{b}{2a}=1.$
Multiply by $2a$ , which is nonzero:
$-b=2a.$
Multiply by $-1$ :
$b=-2a.$
Step 4
Find $a$ from the root product
The $x$ -intercepts are roots, where $f(x)=0$ . If they are $r$ and $s$ , then $f(x)=a(x-r)(x-s)$ has constant term $ars$ . So their product is $c/a$ , and the given product makes:
$\frac{35}{a}=-5.$
Multiply by $a$ :
$35=-5a.$
Divide by $-5$ :
$a=-7.$
Step 5
Complete the rule for $f$
Use $b=-2a$ with $a=-7$ :
$b=-2(-7).$
Type $f(x)=-7x^2-2(-7)x+35$ in Desmos. It stores the completed quadratic function for the evaluation.
Step 6
Evaluate the requested output
Type $f(3)$ on the next line. Desmos shows $14$ , so the value of the quadratic function at $x=3$ is $14$ . Grid in 14.

Question 15·25 Super-Hard SAT Advanced Math Questions·Advanced Math

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