What is the vertices f(x)=x^2-8x+9 ?

The vertices f(x)=x^2-8x+9 is Minimum (4,-7)

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vertices f(x)=x^2-8x+9

vertices

f (x) = x^{2} - 8 x + 9

Minimum (4, - 7)

Solution steps

y = x^{2} - 8 x + 9

The parabola parameters are:

a = 1, b = - 8, c = 9

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} = - 7

Therefore the parabola vertex is

(4, - 7)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (4, - 7)

y = 3 x^{2} + 9 x + 9

f (x) = x^{2} + 10 x + 7

f (x) = x^{2} - 3 x - 18

- 2 x + x (x - 2) - 2 x (x - 1) - (x - 3)

(5 x - 8)^{2}

What is the vertices f(x)=x^2-8x+9 ?
The vertices f(x)=x^2-8x+9 is Minimum (4,-7)