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Popular >

range of 4x^2+8

Solution

range

4 x^{2} + 8

Solution

f (x) \geq 8

+1

Solution steps

Vertex of

4 x^{2} + 8 :

Minimum

(0, 8)

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 = 4,

Vertex

(x_{v}, y_{v}) = (0, 8)

f (x) \geq 8

Popular Examples

domain of sqrt(2-x/(x-2))domain

2 - \frac{x}{x - 2}

domain of f(x)=(x-2)^2(x+1)^3(3x-8)domain

f (x) = (x - 2)^{2} (x + 1)^{3} (3 x - 8)

domain of x^2-1 domain

x^{2} - 1

range of f(x)= 1/(x+1)range

f (x) = \frac{1}{x + 1}

domain of y=(1-6x)/3 domain

y = \frac{1 - 6 x}{3}

Frequently Asked Questions (FAQ)

What is the range of 4x^2+8 ?
The range of 4x^2+8 is f(x)>= 8