What is the vertices y=x^2-9x+20 ?

The vertices y=x^2-9x+20 is Minimum (9/2 ,-1/4)

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vertices y=x^2-9x+20

vertices

y = x^{2} - 9 x + 20

Minimum (\frac{9}{2}, - \frac{1}{4})

Solution steps

y = x^{2} - 9 x + 20

The parabola parameters are:

a = 1, b = - 9, c = 20

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

x_{v} = - \frac{( - 9 )}{2 \cdot 1}

Simplify

x_{v} = \frac{9}{2}

Plug in

x_{v} = \frac{9}{2}

to find the

y_{v}

value

y_{v} = - \frac{1}{4}

Therefore the parabola vertex is

(\frac{9}{2}, - \frac{1}{4})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (\frac{9}{2}, - \frac{1}{4})

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

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

What is the vertices y=x^2-9x+20 ?
The vertices y=x^2-9x+20 is Minimum (9/2 ,-1/4)