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

2 Answers

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

Explanation:

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

Delta = 36 - 4 * 2 * 4 = 4

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

Jun 13, 2018

(-\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})