What is the vertices f(x)=x^2-8x+17 ?

The vertices f(x)=x^2-8x+17 is Minimum (4,1)

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vertices f(x)=x^2-8x+17

vertices

f (x) = x^{2} - 8 x + 17

Minimum (4, 1)

Solution steps

y = x^{2} - 8 x + 17

The parabola parameters are:

a = 1, b = - 8, c = 17

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} = 1

Therefore the parabola vertex is

(4, 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (4, 1)

y = 24000 x^{2} + 900 x - 150

16 - (12 x + 8)^{2}

y^{2} - 8 x - 6 y - 23 = 0

g (x) = (x + 3)^{2} + 5

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

What is the vertices f(x)=x^2-8x+17 ?
The vertices f(x)=x^2-8x+17 is Minimum (4,1)