What is the vertices y=x^2+2x-15 ?

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

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vertices y=x^2+2x-15

vertices

y = 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)

f (x) = - 4 (x + 1)^{2} - 5

x^{2} - 4

2 x^{2} + 3 x - 20

y = x^{2} - 2 x^{1}

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