What is the vertices y=x^2-8x+3 ?

The vertices y=x^2-8x+3 is Minimum (4,-13)

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vertices y=x^2-8x+3

vertices

y = x^{2} - 8 x + 3

Minimum (4, - 13)

Solution steps

y = x^{2} - 8 x + 3

The parabola parameters are:

a = 1, b = - 8, c = 3

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} = - 13

Therefore the parabola vertex is

(4, - 13)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (4, - 13)

(6.6) (0)

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

\frac{x}{x ^{2} + 2}

3 x^{2}

f (x) = 3^{x} + 5