What is the vertices f(x)=x^2-10x+9 ?

The vertices f(x)=x^2-10x+9 is Minimum (5,-16)

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vertices f(x)=x^2-10x+9

vertices

f (x) = x^{2} - 10 x + 9

Minimum (5, - 16)

Solution steps

y = x^{2} - 10 x + 9

The parabola parameters are:

a = 1, b = - 10, c = 9

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

x_{v} = - \frac{( - 10 )}{2 \cdot 1}

Simplify

x_{v} = 5

Plug in

x_{v} = 5

to find the

y_{v}

value

y_{v} = - 16

Therefore the parabola vertex is

(5, - 16)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (5, - 16)

y = - x^{2} - 8

x^{2} - 7 x + 12

y = - x^{2} + 1

f (x) = - 2 x^{2} + 20 x + 48

y = x^{2} - 15 x

What is the vertices f(x)=x^2-10x+9 ?
The vertices f(x)=x^2-10x+9 is Minimum (5,-16)