What is the vertices 5(x+2)^2-3 ?

The vertices 5(x+2)^2-3 is Minimum (-2,-3)

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vertices 5(x+2)^2-3

vertices

5 (x + 2)^{2} - 3

Minimum (- 2, - 3)

Solution steps

y = 5 (x + 2)^{2} - 3

Rewrite

y = 5 (x + 2)^{2} - 3

in the form

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

y = 5 x^{2} + 20 x + 17

The parabola parameters are:

a = 5, b = 20, c = 17

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

x_{v} = - \frac{20}{2 \cdot 5}

Simplify

x_{v} = - 2

Plug in

x_{v} = - 2

to find the

y_{v}

value

y_{v} = - 3

Therefore the parabola vertex is

(- 2, - 3)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 5

Minimum (- 2, - 3)

y^{2} - 8 x = 0

f (x) = - x^{2} + 8 x - 15

y = x^{2} + 6

y = x^{2} + 2

x^{2} + 2 x - 8