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})

y = 3 x^{2} - 12 x + 10

f (x) = - 2 x^{2} - 4 x + 13

y^{2} = - 16 x

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

y = x^{2} - 5 x - 24

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