What is the vertices f(x)=-2(x-3)^2 ?

The vertices f(x)=-2(x-3)^2 is Maximum (3,0)

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vertices f(x)=-2(x-3)^2

vertices

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

Maximum (3, 0)

Solution steps

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

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

The parabola parameters are:

a = - 2, m = 3, n = 3

x_{v} = \frac{m + n}{2}

x_{v} = \frac{3 + 3}{2}

Simplify

x_{v} = 3

Plug in

x_{v} = 3

to find the

y_{v}

value

y_{v} = 0

Therefore the parabola vertex is

(3, 0)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 2

Maximum (3, 0)

y = x^{2} + 4 x + 7

- x^{2} - 4 x + 5

f (x) = x^{2} + 16 x + 64

y = 2 x^{2} - 4 x + 5

y = 6 x^{2} - 1

What is the vertices f(x)=-2(x-3)^2 ?
The vertices f(x)=-2(x-3)^2 is Maximum (3,0)