What is the vertices f(x)=x^2-10x+16 ?

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

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vertices f(x)=x^2-10x+16

vertices

f (x) = x^{2} - 10 x + 16

Minimum (5, - 9)

Solution steps

y = x^{2} - 10 x + 16

The parabola parameters are:

a = 1, b = - 10, c = 16

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} = - 9

Therefore the parabola vertex is

(5, - 9)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (5, - 9)

- 3 x^{2} - 5

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

X^{2} + 2 X - 8

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

f (x) = 2 x^{2} + 6 x + 1

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