What is the vertices 4x^2-20x-24y+97=0 ?

The vertices 4x^2-20x-24y+97=0 is Minimum (5/2 ,3)

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vertices 4x^2-20x-24y+97=0

vertices

4 x^{2} - 20 x - 24 y + 97 = 0

Minimum (\frac{5}{2}, 3)

Solution steps

4 x^{2} - 20 x - 24 y + 97 = 0

Rewrite

4 x^{2} - 20 x - 24 y + 97 = 0

in the form

y = a x^{2} + b x + c

y = \frac{1 \cdot x ^{2}}{6} - \frac{5 x}{6} + \frac{97}{24}

The parabola parameters are:

a = \frac{1}{6}, b = - \frac{5}{6}, c = \frac{97}{24}

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

x_{v} = - \frac{( - \frac{5}{6} )}{2 ( \frac{1}{6} )}

Simplify

- \frac{- \frac{5}{6}}{2 ( \frac{1}{6} )} : \frac{5}{2}

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

Plug in

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

to find the

y_{v}

value

y_{v} = 3

Therefore the parabola vertex is

(\frac{5}{2}, 3)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = \frac{1}{6}

Minimum (\frac{5}{2}, 3)

y - 2 = (x - 1)^{2}

y = (x - 5)^{2}

y = - x^{2} - 4 x - 2

x^{2} - 16

2 x^{2} + 2 x - 4

What is the vertices 4x^2-20x-24y+97=0 ?
The vertices 4x^2-20x-24y+97=0 is Minimum (5/2 ,3)