What is the vertices y=-2x^2+5 ?

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

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vertices y=-2x^2+5

vertices

y = - 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) = - x^{2} - 2 x + 15

- 2.318 + 0.2356 x - 0.002674 x^{2}

y = 4 x^{2} + 5 x - 7

y = x^{2} + x + 8

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