What is the vertices f(x)=x^2-4x-12 ?

The vertices f(x)=x^2-4x-12 is Minimum (2,-16)

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vertices f(x)=x^2-4x-12

vertices

f (x) = x^{2} - 4 x - 12

Minimum (2, - 16)

Solution steps

y = x^{2} - 4 x - 12

The parabola parameters are:

a = 1, b = - 4, c = - 12

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

x_{v} = - \frac{( - 4 )}{2 \cdot 1}

Simplify

x_{v} = 2

Plug in

x_{v} = 2

to find the

y_{v}

value

y_{v} = - 16

Therefore the parabola vertex is

(2, - 16)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (2, - 16)

25 y^{2} - 36 (x - 8)^{2} = 900

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

x = - \frac{1}{2} (y + 3)^{2} + 1

\frac{x ^{2}}{16} - \frac{y ^{2}}{9} = 1

4 x^{2} + 9 y^{2} = 36

What is the vertices f(x)=x^2-4x-12 ?
The vertices f(x)=x^2-4x-12 is Minimum (2,-16)