What is the vertices f(x)=x^2-8x+18 ?

The vertices f(x)=x^2-8x+18 is Minimum (4,2)

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vertices f(x)=x^2-8x+18

vertices

f (x) = x^{2} - 8 x + 18

Minimum (4, 2)

Solution steps

y = x^{2} - 8 x + 18

The parabola parameters are:

a = 1, b = - 8, c = 18

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} = 2

Therefore the parabola vertex is

(4, 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (4, 2)

y^{2} - 4 x - 4 = 0

y = 3 x^{2} + 6 x - 1

(8 x - 23)^{2} - 10

y = (x + 1)^{2} + 1

y = (x + 1)^{2} + 2

What is the vertices f(x)=x^2-8x+18 ?
The vertices f(x)=x^2-8x+18 is Minimum (4,2)