What is the vertices f(x)=(x+4)^2+1 ?

The vertices f(x)=(x+4)^2+1 is Minimum (-4,1)

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vertices f(x)=(x+4)^2+1

vertices

f (x) = (x + 4)^{2} + 1

Minimum (- 4, 1)

Solution steps

y = (x + 4)^{2} + 1

Rewrite

y = (x + 4)^{2} + 1

in the form

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

y = x^{2} + 8 x + 17

The parabola parameters are:

a = 1, b = 8, c = 17

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

x_{v} = - \frac{8}{2 \cdot 1}

Simplify

x_{v} = - 4

Plug in

x_{v} = - 4

to find the

y_{v}

value

y_{v} = 1

Therefore the parabola vertex is

(- 4, 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 4, 1)

f (x) = 2 x^{2} + 8 x + 9

y = - x^{2} + 8 x - 15

f (x) = - 2 x^{2} + 8 x - 5

f (x) = - 2 x^{2} + 8 x - 9

f (x) = - 5 x^{2} + 10 x - 6

What is the vertices f(x)=(x+4)^2+1 ?
The vertices f(x)=(x+4)^2+1 is Minimum (-4,1)