Question 25·SAT Desmos Regression Practice·Advanced Math

The input of $f$ is $3-x$, so make the entire input $3-x$ equal $8$. Set the inputs equal:

Subtract $3$:

Multiply by $-1$:

So $g(-5)$ is tied to $f(8)$. Using $g(8)$ instead would reach $f(-5)$.

The vertex is the turning point of a quadratic graph. With vertex $(5,19)$, the vertex form of $g$ is

At $x=5$, the squared part is $0$, so the output is $19$. The coefficient $a$ is still unknown.

The graph passes through $(3,-5)$, so $g(3)=-5$. Type $-5\sim a(3-5)^2+19$ in Desmos. The regression symbol $\sim$ asks Desmos to find the unknown coefficient that fits this point. Under PARAMETERS, Desmos shows $a=-6$.

Define $g(x)=a(x-5)^2+19$ in Desmos, then type $g(-5)$. Desmos uses the $a$ it found and shows $-581$. That’s $g(-5)$, not yet the value of $f(8)$.

At $x=-5$, the given rule says $g(-5)=-2f(8)+7$. To recover $f(8)$, undo the $+7$ and then the multiplication by $-2$. Subtract $7$:

Divide by $-2$:

Type $\frac{7-g(-5)}{2}$ in Desmos. It shows $294$, so $f(8)=294$. Grid in 294.