Question

How do you find the quadratic function with vertex (-5/2, 0) and point (-7/2, -16/3)?

Answer 1

Because you asked for a quadratic function, we shall discard the vertex form `x=a(y-k)^2+h` and use only:

`y = a(x-h)^2+k" [1]"`

where `(h,k)` is the vertex.

Explanation:

Substitute `-5/2` for h and `0` for k into equation [1]:

`y = a(x--5/2)^2+0`

Simplify:

`y = a(x+5/2)^2" [2]"`

Substitute `-7/2` for x and `-16/3` for y into equation [2]:

`-16/3 = a(-7/2+5/2)^2`

Solve for a:

`-16/3 = a(-2/2)^2 = a(-1)^2`

`a = -16/3`

Substitute `-16/3` for a into equation [2]:

`y = -16/3(x+5/2)^2" [3]"`

`y = -16/3(x^2+5x+ 25/4)`

`y = -16/3x^2-80/3x-100/3" [4]"`

Equation [3] is the vertex form and equation [4] is the standard form.