What is the vertices y=-4x^2+8x+2 ?

The vertices y=-4x^2+8x+2 is Maximum (1,6)

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vertices y=-4x^2+8x+2

vertices

y = - 4 x^{2} + 8 x + 2

Maximum (1, 6)

Solution steps

y = - 4 x^{2} + 8 x + 2

The parabola parameters are:

a = - 4, b = 8, c = 2

x_{v} = - \frac{b}{2 a}

x_{v} = - \frac{8}{2 ( - 4 )}

Simplify

- \frac{8}{2 ( - 4 )} : 1

x_{v} = 1

Plug in

x_{v} = 1

to find the

y_{v}

value

y_{v} = 6

Therefore the parabola vertex is

(1, 6)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 4

Maximum (1, 6)

y^{2} + 12 x - 2 y - 11 = 0

f (x) = x^{2} + 5 x - 2

x^{2} + 6 x + 3 y + 2 = 0

y = - x^{2} + 8 x - 7

- 3 x^{2} + 6 x + 5