Question

What is the vertex form of `y=5x^2+4x+7`?

Answer 1

`y=5(x+2/5)^2+31/5`, where vertex is `(-2/5,31/5)`

Explanation:

Vertex form of equation is of type `y = a(x – h)^2 + k`, where `(h,k)` is the vertex. For this, in the equation `y=5x^2+4x+7`, one should first take `5` out out of first two terms and then make it complete square, as follows:

`y=5x^2+4x+7=5(x^2+4/5x)+7`

To make `(x^2+4/5x)`, complete square, one has to add and subtract, 'square of half the coefficient of `x`, and thus this becomes

`y=5x^2+4x+7=5(x^2+4/5x+(2/5)^2)+7-5*(2/5)^2` or

`y=5(x+2/5)^2+7-4/5` or

`y=5(x-(-2/5))^2+31/5`, where vertex is `(-2/5,31/5)`