What is the vertices (x+5)^2=-20(y-2) ?

The vertices (x+5)^2=-20(y-2) is Maximum (-5,2)

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vertices (x+5)^2=-20(y-2)

vertices

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

Maximum (- 5, 2)

Solution steps

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

Rewrite

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

in the form

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

y = - \frac{x ^{2}}{20} - \frac{x}{2} + \frac{3}{4}

The parabola parameters are:

a = - \frac{1}{20}, b = - \frac{1}{2}, c = \frac{3}{4}

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

x_{v} = - \frac{( - \frac{1}{2} )}{2 ( - \frac{1}{20} )}

Simplify

- \frac{- \frac{1}{2}}{2 ( - \frac{1}{20} )} : - 5

x_{v} = - 5

Plug in

x_{v} = - 5

to find the

y_{v}

value

y_{v} = 2

Therefore the parabola vertex is

(- 5, 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - \frac{1}{20}

Maximum (- 5, 2)

7 x^{2} - 10 x + 5

f (x) = - 2 x^{2} + 80 x - 10

y = 6 x^{2} - 7 x - 3

f (x) = 4 + 12 x - 3 x^{2}

f (x) = - 5 x^{2} - 25 x - 30

What is the vertices (x+5)^2=-20(y-2) ?
The vertices (x+5)^2=-20(y-2) is Maximum (-5,2)