Question 7·Hard·Advanced Math

Let $g(x)=2x+3$ give the line’s height, and let $d(x)=f(x)-g(x)$ be the difference between the two heights. At an intersection, $d(x)=0$. They meet exactly once, at $x=3$, so $d$ is a quadratic with only one real zero: $3$.

A quadratic with one real zero has that zero twice, called a double root. So $d(x)=a(x-3)^2$ for some nonzero number $a$. Subtracting the line reveals symmetry about $x=3$.

$-1$ and $7$ are each four units from $3$. Their squared distances match, so $d(-1)=d(7)$. In terms of the original heights, that says $f(7)-g(7)=f(-1)-g(-1)$. It does not say $f(7)=f(-1)$: the line rises between these inputs.

Add $g(7)$ to both sides, so the change in the line’s height is included: $f(7)=f(-1)+g(7)-g(-1)$.

Type $g(x)=2x+3$, then $g(7)-g(-1)$ in Desmos. It shows $16$. That’s the line’s increase from the known input to the requested input.

Type $19+g(7)-g(-1)$. Desmos shows $35$. The gaps from the line match, so add the line’s increase once to $f(-1)=19$. The value of $f(7)$ is $35$. Choice B.

If the requested input weren’t a mirror of $-1$ across $3$, you could find the whole function. The one-point touch gives $f(x)=a(x-3)^2+2x+3$. The squared term is zero at $x=3$ and nonzero everywhere else because $a\ne0$.

Since $f(-1)=19$, type $19\sim a(-1-3)^2+2(-1)+3$. The $\sim$ tells Desmos to fit the unknown $a$; under PARAMETERS, it shows $a=1.125$. This number sets the size of the squared term.

Type $f(x)=a(x-3)^2+2x+3$, using Desmos’s stored $a$, then type $f(7)$. Desmos shows $35$, the output at the requested input. Choice B.