Question

What is the vertex of `y=2x^2+6x+4`?

Answer 1

`V = (-3/2, - 1/2)`

Explanation:

`V = (-b/(2a), - Delta/(4a))`

`Delta = 36 - 4 * 2 * 4 = 4`

`V = (-6/4, - 4/8)`

Answer 2

`(-\frac{3}{2},-\frac{1}{2})`

Explanation:

Method 1: Calculus approach

Vertex is where the gradient of the curve is 0.

Therefore find `\frac{dy}{dx}`

`\frac{dy}{dx}=4x+6`

Equate that to 0 such that:

`4x+6=0`

Solve for `x`, `x=-\frac{3}{2}`

Let `x=-\frac{3}{2}` into the original function, therefore

`y=2*(-\frac{3}{2})^{2}+6*(-\frac{3}{2})+4`

`y=-\frac{1}{2}`

Method 2: Algebraic approach.

Complete the square to find the turning points, also known as the vertex.

`y=2x^{2}+6x+4`
`y=2(x^{2}+3x+2)`
`y=2[(x+\frac{3}{2})^{2}-\frac{9}{3}+2]`
`y=2(x+\frac{3}{2})^{2}-\frac{1}{2}`

Notice here that you have to multiply BOTH terms by 2, as 2 was the common factor which you took out of the entire expression!

Therefore, the turning points can be picked up such that
`x=-\frac{3}{2},y=-\frac{1}{2}`

Therefore coordinates:

`(-\frac{3}{2},-\frac{1}{2})`