What is the vertices f(x)=x^2-8x ?

The vertices f(x)=x^2-8x is Minimum (4,-16)

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vertices f(x)=x^2-8x

vertices

f (x) = x^{2} - 8 x

Minimum (4, - 16)

Solution steps

y = x^{2} - 8 x

The parabola parameters are:

a = 1, b = - 8, c = 0

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} = - 16

Therefore the parabola vertex is

(4, - 16)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (4, - 16)

x^{2} + 16 y = 0

+ 3 x^{2} + x + 5

y = 3 x^{2} + x - 1

28 x^{2} + 47 x + 15

y^{2} - 8 y + 4 x = 0