Question

What is the the vertex of `y=4x^2 + 9x + 15`?

Answer 1

`y=4(x-(-9/8))^2+159/16`, where vertex is `(-9/8,159/16)`

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=4x^2+9x+15`, one should first take `4` out out of first two terms and then make it complete square, as follows:

`y=4x^2+9x+15=4(x^2+9/4x)+15`

To make `(x^2+9/4x)`, complete square, one has to add and subtract, 'square of half the coefficient of `x`, and thus this becomes

`y=4x^2+9x+15=4(x^2+9/4x+(9/8)^2)+15-4*(9/8)^2` or

`y=4(x+9/8)^2+15-81/16` or

`y=4(x-(-9/8))^2+159/16`, where vertex is `(-9/8,159/16)`