What is the vertices y=-2(x-3)^2+4 ?

The vertices y=-2(x-3)^2+4 is Maximum (3,4)

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vertices y=-2(x-3)^2+4

vertices

y = - 2 (x - 3)^{2} + 4

Maximum (3, 4)

Solution steps

y = - 2 (x - 3)^{2} + 4

Rewrite

y = - 2 (x - 3)^{2} + 4

in the form

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

y = - 2 x^{2} + 12 x - 14

The parabola parameters are:

a = - 2, b = 12, c = - 14

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

x_{v} = - \frac{12}{2 ( - 2 )}

Simplify

- \frac{12}{2 ( - 2 )} : 3

x_{v} = 3

Plug in

x_{v} = 3

to find the

y_{v}

value

y_{v} = 4

Therefore the parabola vertex is

(3, 4)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 2

Maximum (3, 4)

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

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

f (x) = x^{2} - 10 x + 25

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

- x^{2} + 10 x

What is the vertices y=-2(x-3)^2+4 ?
The vertices y=-2(x-3)^2+4 is Maximum (3,4)