Question

Find the equation of the parabola which has the given vertex V, which passes through the given point P, and which has the specified axis of symmetry. V(4,−2),P(2,14), vertical axis of symmetry.

Answer 1

`y = 4(x - 4)^2 - 2`

or

`y=4x^2 -32x + 62`

Explanation:

Given

`V = (4,-2)` --- vertex

`P = (2,14)` --- point

Required

The equation of the parabola

The equation of the parabola is of the form

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

Where

`V (4,-2) = (h,k)` ---- the vertex

So, we have:

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

`y = a(x - 4)^2 - 2`

In
`P = (2,14)`, we have:

`(x,y) = (2,14)`

Substitute
`(x,y) = (2,14)` in
`y = a(x - 4)^2 - 2`

`14 = a(2 - 4)^2 - 2`

`14 = a(- 2)^2 - 2`

`14 = a*4 - 2`

`14 = 4a - 2`

Collect like terms

`4a = 14 +2`

`4a = 16`

Divide both sides by 4

`a= 4`

So:

`y = a(x - 4)^2 - 2` becomes

`y = 4(x - 4)^2 - 2`

Open bracket to express the equation in standard form

`y=4(x^2 -8x + 16) - 2`

`y=4x^2 -32x + 64 - 2`

`y=4x^2 -32x + 62`