What is the vertices f(x)=4(x-2)^2+11 ?

The vertices f(x)=4(x-2)^2+11 is Minimum (2,11)

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vertices f(x)=4(x-2)^2+11

vertices

f (x) = 4 (x - 2)^{2} + 11

Minimum (2, 11)

Solution steps

y = 4 (x - 2)^{2} + 11

Rewrite

y = 4 (x - 2)^{2} + 11

in the form

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

y = 4 x^{2} - 16 x + 27

The parabola parameters are:

a = 4, b = - 16, c = 27

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} = 11

Therefore the parabola vertex is

(2, 11)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 4

Minimum (2, 11)

f (x) = (2 x + 2) (x - 5)

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

y = 4 - (x - 3)^{2}

f (x) = 2 x^{2} - 10 x

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

What is the vertices f(x)=4(x-2)^2+11 ?
The vertices f(x)=4(x-2)^2+11 is Minimum (2,11)