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

The vertices f(x)=x^2-2x-8 is Minimum (1,-9)

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

vertices

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

Minimum (1, - 9)

Solution steps

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

The parabola parameters are:

a = 1, b = - 2, c = - 8

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} = - 9

Therefore the parabola vertex is

(1, - 9)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (1, - 9)

x^{2} - y = 4

16 x^{2} + y^{2} = 16

y = x^{2} - 6 x

x^{2} + 4 y^{2} - 6 x - 16 y + 21 = 0

\frac{x ^{2}}{25} - \frac{y ^{2}}{16} = 1

What is the vertices f(x)=x^2-2x-8 ?
The vertices f(x)=x^2-2x-8 is Minimum (1,-9)