What is the vertices f(x)=-x^2-2x+15 ?

The vertices f(x)=-x^2-2x+15 is Maximum (-1,16)

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vertices f(x)=-x^2-2x+15

vertices

f (x) = - x^{2} - 2 x + 15

Maximum (- 1, 16)

Solution steps

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

The parabola parameters are:

a = - 1, b = - 2, c = 15

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

x_{v} = - \frac{( - 2 )}{2 ( - 1 )}

Simplify

x_{v} = - 1

Plug in

x_{v} = - 1

to find the

y_{v}

value

y_{v} = 16

Therefore the parabola vertex is

(- 1, 16)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (- 1, 16)

- 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}

2 x^{2} - 4 x + 7

What is the vertices f(x)=-x^2-2x+15 ?
The vertices f(x)=-x^2-2x+15 is Maximum (-1,16)