Question

1. write the standard form of the quadratic with the vertex (4, 1) and goes through (-4, 33)

Explanation:

Step1: Recall vertex - form of quadratic

The vertex - form of a quadratic function is $y=a(x - h)^2+k$, where $(h,k)$ is the vertex. Given the vertex $(4,1)$, the equation is $y=a(x - 4)^2+1$.

Step2: Find the value of $a$

Since the quadratic goes through the point $(-4,33)$, substitute $x=-4$ and $y = 33$ into the equation $y=a(x - 4)^2+1$. We get $33=a(-4 - 4)^2+1$. First, simplify the right - hand side: $(-4 - 4)^2=(-8)^2 = 64$. So the equation becomes $33=64a+1$. Subtract 1 from both sides: $33−1=64a$, which gives $32 = 64a$. Then divide both sides by 64: $a=\frac{32}{64}=\frac{1}{2}$.

Step3: Write the standard form

Substitute $a=\frac{1}{2}$ back into the vertex - form $y=a(x - 4)^2+1$.
\[

$$\begin{align*} y&=\frac{1}{2}(x - 4)^2+1\\ &=\frac{1}{2}(x^{2}-8x + 16)+1\\ &=\frac{1}{2}x^{2}-4x+8 + 1\\ &=\frac{1}{2}x^{2}-4x + 9 \end{align*}$$

\]

Answer:

$y=\frac{1}{2}x^{2}-4x + 9$