What is the vertices f(x)=x^2-6x+5 ?

The vertices f(x)=x^2-6x+5 is Minimum (3,-4)

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vertices f(x)=x^2-6x+5

vertices

f (x) = x^{2} - 6 x + 5

Minimum (3, - 4)

Solution steps

y = x^{2} - 6 x + 5

The parabola parameters are:

a = 1, b = - 6, c = 5

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

x_{v} = - \frac{( - 6 )}{2 \cdot 1}

Simplify

x_{v} = 3

Plug in

x_{v} = 3

to find the

y_{v}

value

y_{v} = - 4

Therefore the parabola vertex is

(3, - 4)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (3, - 4)

\frac{x ^{2}}{16} + \frac{y ^{2}}{7} = 1

x^{2} + y = 1

x = y^{2} - 3

x^{2} - y^{2} = 49

\frac{y ^{2}}{49} - \frac{x ^{2}}{25} = 1

What is the vertices f(x)=x^2-6x+5 ?
The vertices f(x)=x^2-6x+5 is Minimum (3,-4)