What is the vertices y=x^2+8x+7 ?

The vertices y=x^2+8x+7 is Minimum (-4,-9)

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vertices y=x^2+8x+7

vertices

y = x^{2} + 8 x + 7

Minimum (- 4, - 9)

Solution steps

y = x^{2} + 8 x + 7

The parabola parameters are:

a = 1, b = 8, c = 7

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

x_{v} = - \frac{8}{2 \cdot 1}

Simplify

x_{v} = - 4

Plug in

x_{v} = - 4

to find the

y_{v}

value

y_{v} = - 9

Therefore the parabola vertex is

(- 4, - 9)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 4, - 9)

4 t^{2} + 1

f (x) = \frac{1}{2} (x - 1)^{2}

x^{2} + 1

2 x^{2} - 5 x + 3

2 X^{2} + 3 X - 20