What is the vertices y=x^2-11x+13 ?

The vertices y=x^2-11x+13 is Minimum (11/2 ,-69/4)

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vertices y=x^2-11x+13

vertices

y = x^{2} - 11 x + 13

Minimum (\frac{11}{2}, - \frac{69}{4})

Solution steps

y = x^{2} - 11 x + 13

The parabola parameters are:

a = 1, b = - 11, c = 13

x_{v} = - \frac{b}{2 a}

x_{v} = - \frac{( - 11 )}{2 \cdot 1}

Simplify

x_{v} = \frac{11}{2}

Plug in

x_{v} = \frac{11}{2}

to find the

y_{v}

value

y_{v} = - \frac{69}{4}

Therefore the parabola vertex is

(\frac{11}{2}, - \frac{69}{4})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (\frac{11}{2}, - \frac{69}{4})

What is the vertices y=x^2-11x+13 ?
The vertices y=x^2-11x+13 is Minimum (11/2 ,-69/4)