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

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

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

vertices

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

Minimum (- 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 \cdot 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

Minimum (- 1, - 16)

8 y = x^{2}

x = y^{2} + 3 y + 4

y = 2 x^{2} - 5

y = 500 (0.04 - x^{2})

- 5 x^{2} - 40 x - 77

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