Question 45·200 Super-Hard SAT Math Questions·Advanced Math

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Function $f$ is quadratic and $f(0)=11$ . For each value of $x$ , a function $g$ is defined by $g(x)=f(x+3)-f(x-1)$ . The graph of $y=g(x)$ has a $y$ -intercept of $(0,-16)$ and an $x$ -intercept of $(3,0)$ . What is the value of $f(6)$ ?

Your answer

A quadratic, a function with an $x^2$ term, has a known constant term when its output at $0$ is given. If a second function subtracts shifted outputs of that quadratic, turn its intercepts into equations about the original function. Use a Desmos list regression to find the missing coefficients, then evaluate the original function. An intercept of the difference is not necessarily an intercept of the quadratic.

Hints

Hint 1
In a quadratic written as $f(x)=ax^2+bx+c$ , putting in $0$ makes both terms containing $x$ vanish. Use the given output at $0$ to find $c$ first.
Hint 2
An intercept of $g$ describes $g$ , not $f$ . At a $y$ -intercept the input is $0$ ; at an $x$ -intercept the output is $0$ . What happens when you replace $x$ in both shifted inputs of $g$ ?
Hint 3
Subtracting two outputs of the same quadratic cancels its constant term. That leaves two equations in $a$ and $b$ , which you can enter together as a Desmos regression, a fit for unknown numbers.

Step-by-step

Turn the intercepts into a Desmos regression
Step 1
Write the original quadratic
A quadratic has the form $f(x)=ax^2+bx+c$ . At $x=0$ , the terms containing $x$ disappear, so $f(0)=c$ . The given $f(0)=11$ makes the constant term $11$ :
$f(x)=ax^2+bx+11.$
Step 2
Translate the y-intercept of g
The $y$ -intercept $(0,-16)$ says $g(0)=-16$ . Replace $x$ in both inputs of $g(x)=f(x+3)-f(x-1)$ :
$g(0)=f(3)-f(-1)=-16.$
An intercept of $g$ gives an equation about outputs of $f$ ; it is not an intercept of $f$ .
Step 3
Translate the x-intercept of g
At the $x$ -intercept $(3,0)$ , the output of $g$ is $0$ . Put $3$ in for $x$ in both inputs:
$g(3)=f(6)-f(2)=0.$
This says $f(6)=f(2)$ , not that $f(6)=0$ .
Step 4
Express the first difference using a and b
Substitute $3$ and $-1$ into the same quadratic, then subtract:
$-16=(9a+3b+11)-(a-b+11).$
Distribute the minus sign across the whole second output:
$-16=9a+3b+11-a+b-11.$
Combine like terms:
$-16=8a+4b.$
Step 5
Express the second difference using a and b
Substitute $6$ and $2$ into $f$ , then use the zero output of $g$ :
$0=(36a+6b+11)-(4a+2b+11).$
Distribute the minus sign:
$0=36a+6b+11-4a-2b-11.$
Combine like terms:
$0=32a+4b.$
Step 6
Fit both coefficients at once
Type $[8a+4b,32a+4b]\sim[-16,0]$ in Desmos. The $\sim$ tells a regression to find $a$ and $b$ that match both equations. Under PARAMETERS, Desmos shows $a\approx0.6667$ and $b\approx-5.3333$ . These are coefficients of $f$ , not the requested function value.
Step 7
Define f with the fitted coefficients
Type $f(x)=ax^2+bx+11$ below the regression. Desmos uses the stored values of $a$ and $b$ , rather than the rounded decimals displayed under PARAMETERS. You now have the original quadratic ready to evaluate.
Step 8
Evaluate the requested output
Type $f(6)$ . Desmos prints $3$ , the output of $f$ at input $6$ . That is the value the question asks you to enter. Grid in 3.

Question 45·200 Super-Hard SAT Math Questions·Advanced Math

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