Question 85·Medium·Advanced Math

The graph labels $(2,-8)$ as the vertex, the point where the parabola turns. In vertex form, the square becomes $0$ at the vertex's $x$-coordinate, so $(2,-8)$ gives $y=a(x-2)^2-8$. The vertex sets the location, but it doesn't tell you how steep the curve is; that's what $a$ controls.

The marked point $(0,4)$ says that $y=4$ when $x=0$. Put those coordinates into the vertex form: $4=a(0-2)^2-8$. Don't use the vertex again to find $a$: its squared term would be $0$, no matter what $a$ is.

Type $4\sim a(0-2)^2-8$. The regression symbol $\sim$ asks Desmos to find the unknown $a$ that makes the point's equation true. Under PARAMETERS, Desmos shows $a=3$. That's the number multiplying the square, not the requested $b$.

Expand the squared part: $(x-2)^2=x^2-4x+4$. The $-4x$ comes from multiplying $x$ by $-2$ twice. That's the term you'll need to compare with $bx$.

In $y=a(x-2)^2-8$, the $a=3$ multiplies that $-4x$ term. Multiply it: $3(-4x)=-12x$. A coefficient is the number multiplying a variable, so in the given $y=ax^2+bx+c$, $b=-12$. The point $(0,4)$ gives $c=4$, not $b$. Grid in -12.