What is the vertices y=x^2-8x+17 ?

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

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vertices y=x^2-8x+17

vertices

y = 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 = 2 x^{2} - 4 x + 15

4 (x + 6) + 3 (2 + x + 3 x^{2}) - 2 x^{2}

f (x) = - x^{2} + 6 x - 11

y^{2} = - 20 x

2 x^{2} - 6 x + 15