What is the vertices f(x)=x^2-10x+21 ?

The vertices f(x)=x^2-10x+21 is Minimum (5,-4)

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vertices f(x)=x^2-10x+21

vertices

f (x) = x^{2} - 10 x + 21

Minimum (5, - 4)

Solution steps

y = x^{2} - 10 x + 21

The parabola parameters are:

a = 1, b = - 10, c = 21

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} = - 4

Therefore the parabola vertex is

(5, - 4)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (5, - 4)

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

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

(x - 1)^{2} = - 4 (y - 2)

f (x) = \frac{1}{3} x^{2} + 2 x + 1

4 x^{2} - 8 x + 2 y = 5

What is the vertices f(x)=x^2-10x+21 ?
The vertices f(x)=x^2-10x+21 is Minimum (5,-4)