What is the vertices f(x)=-2x^2+5 ?

The vertices f(x)=-2x^2+5 is Maximum (0,5)

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vertices f(x)=-2x^2+5

vertices

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

Maximum (0, 5)

Solution steps

y = - 2 x^{2} + 5

The parabola parameters are:

a = - 2, b = 0, c = 5

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

x_{v} = - \frac{0}{2 ( - 2 )}

Simplify

x_{v} = 0

Plug in

x_{v} = 0

to find the

y_{v}

value

y_{v} = 5

Therefore the parabola vertex is

(0, 5)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 2

Maximum (0, 5)

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

2 x^{2} - 12 x + 3

y = - 3 x^{2} + 6 x - 1

y = (x + 5)^{2}

y = x^{2} + 5 x - 14