Question

What is the vertex form of `5y = 13x^2+20x+42`?

Answer 1

`y = 13/5(x - -10/13)^2 + 446/65`

Explanation:

Divide both sides by 5:

`y = 13/5x^2 + 4x + 42/5`

The equation is in standard form, `y = ax^2 + bx + c`. In this form the x coordinate, h, of the vertex is:

`h = -b/(2a)`

`h = - 4/(2(13/5)) = -20/26 = -10/13`

The y coordinate, k, of the vertex is the function evaluated at h.

`k = 13/5(-10/13)^2 + 4(-10/13) + 42/5`

`k = 13/5(-10/13)(-10/13) - 40/13 + 42/5`

`k = (-2)(-10/13) - 40/13 + 42/5`

`k = 20/13 - 40/13 + 42/5`

`k = -20/13 + 42/5`

`k = -100/65 + 546/65`

`k = 446/65`

The vertex form of the equation of a parabola is:

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

Substituting in our known values:

`y = 13/5(x - -10/13)^2 + 446/65`