Question

What is the standard form of the quadratic function that has a vertex at (4, 5) and goes through the point (5, 6)?

Answer 1

The vertex form of the equation of a quadratic function is given to be:

$y=a(x-h)^2+k$

where (h, k) is the vertex.

From the question, we have the following parameters:

$\begin{gathered} (h,k)=(4,5) \\ (x,y)=(5,6) \end{gathered}$

Therefore, the equation will be in the form:

$y=a(x-4)^2+5$

At the point (5, 6), we can get the value of a to be:

$\begin{gathered} 6=a(5-4)^2+5 \\ a=1 \end{gathered}$

Therefore, we have the vertex form to be:

$y=(x-4)^2+5$

Expanding, we have the equation to be:

$\begin{gathered} y=x^2-8x+16+5 \\ y=x^2-8x+21 \end{gathered}$

OPTION B is correct.