What is the vertices f(x)=4(x+4)^2+6 ?

The vertices f(x)=4(x+4)^2+6 is Minimum (-4,6)

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vertices f(x)=4(x+4)^2+6

vertices

f (x) = 4 (x + 4)^{2} + 6

Minimum (- 4, 6)

Solution steps

y = 4 (x + 4)^{2} + 6

Rewrite

y = 4 (x + 4)^{2} + 6

in the form

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

y = 4 x^{2} + 32 x + 70

The parabola parameters are:

a = 4, b = 32, c = 70

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

x_{v} = - \frac{32}{2 \cdot 4}

Simplify

x_{v} = - 4

Plug in

x_{v} = - 4

to find the

y_{v}

value

y_{v} = 6

Therefore the parabola vertex is

(- 4, 6)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 4

Minimum (- 4, 6)

4 (x + 2)^{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

What is the vertices f(x)=4(x+4)^2+6 ?
The vertices f(x)=4(x+4)^2+6 is Minimum (-4,6)