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Popular >

range of f(x)=8x^2+1

Solution

range

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

Solution

f (x) \geq 1

+1

Interval Notation

[1, \infty)

Solution steps

Vertex of

8 x^{2} + 1 :

Minimum

(0, 1)

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 = 8,

Vertex

(x_{v}, y_{v}) = (0, 1)

f (x) \geq 1

Popular Examples

slope of 3x-5y=15 slope

3 x - 5 y = 15

inverse of f(x)=-5cos(2x)inverse

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

inverse of f(x)= 2/3 w-8 inverse

f (x) = \frac{2}{3} w - 8

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

shift 4cos(2x+pi)shift

4 cos (2 x + π)

Frequently Asked Questions (FAQ)

What is the range of f(x)=8x^2+1 ?
The range of f(x)=8x^2+1 is f(x)>= 1