What is the vertices 4(x+2)^2+5 ?

The vertices 4(x+2)^2+5 is Minimum (-2,5)

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vertices 4(x+2)^2+5

vertices

4 (x + 2)^{2} + 5

Minimum (- 2, 5)

Solution steps

y = 4 (x + 2)^{2} + 5

Rewrite

y = 4 (x + 2)^{2} + 5

in the form

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

y = 4 x^{2} + 16 x + 21

The parabola parameters are:

a = 4, b = 16, c = 21

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

x_{v} = - \frac{16}{2 \cdot 4}

Simplify

x_{v} = - 2

Plug in

x_{v} = - 2

to find the

y_{v}

value

y_{v} = 5

Therefore the parabola vertex is

(- 2, 5)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 4

Minimum (- 2, 5)

x^{2} - 2 x - 80

y^{2} + 2 y - 4 x + 9 = 0

f (x) = 2 x^{2} + 5 x - 12

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

(y + 6)^{2} = 12 (x + 1)