What is the vertices y=-1/4 x^2+8x-4 ?

The vertices y=-1/4 x^2+8x-4 is Maximum (16,60)

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vertices y=-1/4 x^2+8x-4

vertices

y = - \frac{1}{4} x^{2} + 8 x - 4

Maximum (16, 60)

Solution steps

y = - \frac{1}{4} x^{2} + 8 x - 4

Rewrite

y = - \frac{1}{4} x^{2} + 8 x - 4

in the form

y = a x^{2} + b x + c

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

The parabola parameters are:

a = - \frac{1}{4}, b = 8, c = - 4

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

x_{v} = - \frac{8}{2 ( - \frac{1}{4} )}

Simplify

- \frac{8}{2 ( - \frac{1}{4} )} : 16

x_{v} = 16

Plug in

x_{v} = 16

to find the

y_{v}

value

y_{v} = 60

Therefore the parabola vertex is

(16, 60)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - \frac{1}{4}

Maximum (16, 60)

f (x) = - x^{2} + 10 x

- \frac{x ^{2}}{2} - 2 x

y = - x^{2} + 5

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

f (x) = 3 x^{2} - 2 x - 1

What is the vertices y=-1/4 x^2+8x-4 ?
The vertices y=-1/4 x^2+8x-4 is Maximum (16,60)