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)

f (x) = - x^{2} + 7

y = - 4 x^{2} + 8 x + 2

y^{2} + 12 x - 2 y - 11 = 0

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

x^{2} + 6 x + 3 y + 2 = 0