What is the vertices f(x)=x^2-8x+12 ?

The vertices f(x)=x^2-8x+12 is Minimum (4,-4)

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vertices f(x)=x^2-8x+12

vertices

f (x) = x^{2} - 8 x + 12

Minimum (4, - 4)

Solution steps

y = x^{2} - 8 x + 12

The parabola parameters are:

a = 1, b = - 8, c = 12

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} = - 4

Therefore the parabola vertex is

(4, - 4)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (4, - 4)

\frac{y ^{2}}{36} - \frac{x ^{2}}{64} = 1

\frac{( x - 3 ) ^{2}}{5} + \frac{( y - 1 ) ^{2}}{15} = 1

3 x^{2} - y^{2} = 11

9 x^{2} + 16 y^{2} = 144

x^{2} + y^{2} = 10

What is the vertices f(x)=x^2-8x+12 ?
The vertices f(x)=x^2-8x+12 is Minimum (4,-4)