quadratic equation

Question

write an quadratic equation in standard form that has a vertex at (4,8) and passes through the point (3'6)

Answer 1

Hello Peyton,

To solve this question, it is easiest to write the equation in vertex form and the convert to standard form. The general equation for a quadratic equation in vertex form is as follows:

y = A ( x - h) ^2 + k

Where the vertex is at (h,k), and A is a number that we will have to mathematically find. First, we plug in the vertex into our equation. Remember that (h,k) is our vertex, so in this case it is (4,8), so h= 4 and k = 8.

y = A (x - 4)^2 + 8

From here we plug in the x,y point that was given to us and solve for A.

6 = A (3 -4)^2 + 8

6 = A ( -1)^2 + 8

6 = A + 8

A = -2

We rewrite our equation as follows:

y = -2 (x - 4) ^2 + 8

This is in vertex form, but we need it in standard from. To do this, we just expand the parentheses. As a reminder, standard form has the general equation y = ax^2 + bx + c

y = -2 ( x^2 - 8x + 16) + 8

y = -2x^2 + 16x - 32 + 8

y = -2x^2 + 16x - 24

Hope this helps, and good luck!