What is the vertices y=-((x+3)^2)/(20)+1 ?

The vertices y=-((x+3)^2)/(20)+1 is Maximum (-3,1)

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vertices y=-((x+3)^2)/(20)+1

vertices

y = - \frac{( x + 3 ) ^{2}}{20} + 1

Maximum (- 3, 1)

Solution steps

y = - \frac{( x + 3 ) ^{2}}{20} + 1

y = - \frac{( x + 3 ) ^{2}}{20} + 1

The parabola parameters are:

a = - \frac{1}{20}, m = 25 - 3, n = - 25 - 3

x_{v} = \frac{m + n}{2}

x_{v} = \frac{( 2 5 - 3 ) + ( - 2 5 - 3 )}{2}

Simplify

\frac{( 2 5 - 3 ) + ( - 2 5 - 3 )}{2} : - 3

x_{v} = - 3

Plug in

x_{v} = - 3

to find the

y_{v}

value

y_{v} = 1

Therefore the parabola vertex is

(- 3, 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - \frac{1}{20}

Maximum (- 3, 1)

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

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

y = x^{2} - 5 x 3

f (x) = 2 x^{2} + 9 x + 10

f (x) = x^{2} + 6 x - 16

What is the vertices y=-((x+3)^2)/(20)+1 ?
The vertices y=-((x+3)^2)/(20)+1 is Maximum (-3,1)