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

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

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vertices y=-4(x-2)^2+4

vertices

y = - 4 (x - 2)^{2} + 4

Maximum (2, 4)

Solution steps

y = - 4 (x - 2)^{2} + 4

Rewrite

y = - 4 (x - 2)^{2} + 4

in the form

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

y = - 4 x^{2} + 16 x - 12

The parabola parameters are:

a = - 4, b = 16, c = - 12

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

x_{v} = - \frac{16}{2 ( - 4 )}

Simplify

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

x_{v} = 2

Plug in

x_{v} = 2

to find the

y_{v}

value

y_{v} = 4

Therefore the parabola vertex is

(2, 4)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 4

Maximum (2, 4)

4 x - y^{2} + y - \frac{31}{4} = 0

f (x) = x^{2} - 10 x + 16

- 3 x^{2} - 5

f (x) = x^{2} - 8 x - 9

X^{2} + 2 X - 8

What is the vertices y=-4(x-2)^2+4 ?
The vertices y=-4(x-2)^2+4 is Maximum (2,4)