What is the vertices y=x^2+2x-35 ?

The vertices y=x^2+2x-35 is Minimum (-1,-36)

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vertices y=x^2+2x-35

vertices

y = x^{2} + 2 x - 35

Minimum (- 1, - 36)

Solution steps

y = x^{2} + 2 x - 35

The parabola parameters are:

a = 1, b = 2, c = - 35

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} = - 36

Therefore the parabola vertex is

(- 1, - 36)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 1, - 36)

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

f (x) = x^{2} - 10 x + 9

y = - x^{2} - 8

x^{2} - 7 x + 12

y = - x^{2} + 1