Question

The quadratic passes thru the point (-5,8) and the axis of symmetry is x = 3. How do I determine the equation of the quadratic?

Answer 1

These conditions are satisfied by any quadratic of the form:

`f(x) = a(x-3)^2 + 8-64a=ax^2-6ax+(8-55a)`

Explanation:

Since the axis of symmetry is `x=3`, the quadratic can be written in the form:

`f(x) = a(x-3)^2 + b`

Since the quadratic passes through `(-5, 8)` we have:

`8 = f(-5) = a(-5-3)^2+b = 64a+b`

Subtract `64a` from both ends to get:

`b = 8-64a`

Then:

`f(x) = a(x-3)^2 + 8-64a`

`=ax^2-6ax+9a+8-64a`

`=ax^2-6ax+(8-55a)`

Here are some of the quadratics that satisfy the conditions:

graph{(x^2-6x-47-y)(1/4x^2-3/2x+8-55/4-y)(-x^2/10+3x/5+13.5-y)=0 [-32.74, 31.35, -11.24, 20.84]}