What is the vertices x^2-5x-36 ?

The vertices x^2-5x-36 is Minimum (5/2 ,-169/4)

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vertices x^2-5x-36

vertices

x^{2} - 5 x - 36

Minimum (\frac{5}{2}, - \frac{169}{4})

Solution steps

y = x^{2} - 5 x - 36

The parabola parameters are:

a = 1, b = - 5, c = - 36

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

x_{v} = - \frac{( - 5 )}{2 \cdot 1}

Simplify

x_{v} = \frac{5}{2}

Plug in

x_{v} = \frac{5}{2}

to find the

y_{v}

value

y_{v} = - \frac{169}{4}

Therefore the parabola vertex is

(\frac{5}{2}, - \frac{169}{4})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (\frac{5}{2}, - \frac{169}{4})

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

100 (1 - \frac{x}{40})^{2}

y = x^{2} + 12 x + 16

f (x) = - 5 x^{2} + 20 x + 60

f (x) = 2 - x^{2}

What is the vertices x^2-5x-36 ?
The vertices x^2-5x-36 is Minimum (5/2 ,-169/4)