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Popular >

range of f(x)=x^2-8x+16

Solution

range

f (x) = x^{2} - 8 x + 16

Solution

f (x) \geq 0

+1

Interval Notation

[0, \infty)

Solution steps

Vertex of

x^{2} - 8 x + 16 :

Minimum

(4, 0)

For a parabola

a x^{2} + b x + c

with Vertex

(x_{v}, y_{v})

If

a < 0

the range is

f (x) \leq y_{v}

If

a > 0

the range is

f (x) \geq y_{v}

a = 1,

Vertex

(x_{v}, y_{v}) = (4, 0)

f (x) \geq 0

Popular Examples

domain of f(x)= 4/(x-3)domain

f (x) = \frac{4}{x - 3}

inflection f(x)=x^3-x inflection points

f (x) = x^{3} - x

domain of f(x)=((x^2+4))/(x^2+4x-5)domain

f (x) = \frac{( x ^{2} + 4 )}{x ^{2} + 4 x - 5}

domain of sqrt(x+6)domain

x + 6

distance (2,2),(1,1)distance

(2, 2), (1, 1)

Frequently Asked Questions (FAQ)

What is the range of f(x)=x^2-8x+16 ?
The range of f(x)=x^2-8x+16 is f(x)>= 0