Question 192·200 Super-Hard SAT Math Questions·Advanced Math

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Quadratic function $f$ satisfies the equation

f(x) - f(-x) = 24x

for all real numbers $x$ . If $f(2)=5$ and $f(-1)=-4$ , which choice is the maximum value of $f(x)$ ?

An identity is an equation that holds for every input. When a quadratic is compared with its value at $-x$ , the squared and constant terms cancel, revealing the $x$ coefficient. Use the two given function values to fit the remaining coefficients with a Desmos regression, then click the vertex, the parabola’s turning point. A given function value need not be its maximum.

Hints

Hint 1
Write a quadratic as $f(x)=ax^2+bx+c$ . Replacing $x$ with $-x$ changes the sign of only the $bx$ term. What remains when you subtract the two function values?
Hint 2
Because the equation holds for every $x$ , the coefficient of $x$ on each side must match. Once you know $b$ , substitute the two given inputs to make two equations for $a$ and $c$ .
Hint 3
Fit those two equations together in one Desmos list regression. Then graph the resulting quadratic: if its $x^2$ coefficient is negative, the vertex gives the maximum.

Step-by-step

Approach 1: Fit the quadratic, then click its vertex
Step 1
Write the function at the opposite input
A quadratic has the form $f(x)=ax^2+bx+c$ . Substitute $-x$ for $x$ :
$f(-x)=a(-x)^2+b(-x)+c.$
Simplify the signs:
$f(-x)=ax^2-bx+c.$
Step 2
Use the identity to find the $x$ coefficient
Subtract the two expressions:
$f(x)-f(-x)=(ax^2+bx+c)-(ax^2-bx+c)$ $f(x)-f(-x)=2bx.$
The squared and constant terms cancel; only the $x$ term remains. The equation holds for every $x$ , so its $x$ coefficients must match:
$2b=24.$
Divide by $2$ :
$b=12.$
Step 3
Turn both function values into equations
Now $f(x)=ax^2+12x+c$ . Use the input $2$ and its output $5$ :
$a(2)^2+12(2)+c=5$ $4a+24+c=5.$
Use the input $-1$ and its output $-4$ ; keep the negative sign in the $x$ term:
$a(-1)^2+12(-1)+c=-4$ $a-12+c=-4.$
Step 4
Fit the two equations together
Type $[4a+24+c,a-12+c]\sim[5,-4]$ . The regression symbol $\sim$ tells Desmos to find one pair of coefficients that fits both equations. Under PARAMETERS, it shows $a=-9$ and $c=17$ . Don’t stop at $17$ : that is $f(0)$ , not necessarily the maximum.
Step 5
Read the maximum from the vertex
Add $f(x)=ax^2+12x+c$ and click the top of its graph. Desmos labels the vertex about $(0.667,21)$ . Since $a=-9$ is negative, the parabola opens downward, so no output is higher than the vertex’s output. The maximum value of $f(x)$ is $21$ . Choice A.

Approach 2: Show why the vertex is the maximum
Step 1
Factor the squared-term coefficient
The fitted coefficients give $f(x)=-9x^2+12x+17$ . To make its vertex form visible, factor $-9$ from the terms containing $x$ :
$f(x)=-9\left(x^2-\frac43x\right)+17.$
Step 2
Complete the square without changing the function
Half of $-\frac43$ is $-\frac23$ , and its square is $\frac49$ . Adding $\frac49$ inside the parentheses changes the function by $-9\left(\frac49\right)=-4$ , so add $4$ outside to balance it:
$f(x)=-9\left(x^2-\frac43x+\frac49\right)+17+4.$
Rewrite the squared expression and combine the constants:
$f(x)=-9\left(x-\frac23\right)^2+21.$
Step 3
Read the highest possible output
A square cannot be negative, so $-9\left(x-\frac23\right)^2$ cannot be positive. It is $0$ at $x=\frac23$ . That makes $21$ the maximum value of $f(x)$ . Choice A.

Question 192·200 Super-Hard SAT Math Questions·Advanced Math

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