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Popular >

range of f(x)=2x^2+5

Solution

range

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

Solution

f (x) \geq 5

+1

Solution steps

Vertex of

2 x^{2} + 5 :

Minimum

(0, 5)

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 = 2,

Vertex

(x_{v}, y_{v}) = (0, 5)

f (x) \geq 5

Popular Examples

inverse of 3x^2-2 inverse

3 x^{2} - 2

midpoint (2,-2),(5,1)midpoint

(2, - 2), (5, 1)

domain of (4x+4)/(x^2+3x+2)domain

\frac{4 x + 4}{x ^{2} + 3 x + 2}

amplitude of f(x)=0.5cos(6x)amplitude

f (x) = 0.5 cos (6 x)

range of f(x)=sqrt(x^2-3x)range

f (x) = x^{2} - 3 x

Frequently Asked Questions (FAQ)

What is the range of f(x)=2x^2+5 ?
The range of f(x)=2x^2+5 is f(x)>= 5