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Popular >

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

Solution

range

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

Solution

f (x) \geq - 56

+1

Interval Notation

[- 56, \infty)

Solution steps

Vertex of

x^{2} + 16 x + 8 :

Minimum

(- 8, - 56)

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}) = (- 8, - 56)

f (x) \geq - 56

Popular Examples

y = 1

inverse of f(x)=-2x+3 inverse

f (x) = - 2 x + 3

extreme f(x)=5cos^2(x)extreme points

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

critical f(x)=x-e^x critical points

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

range of f(x)=x^3+1 range

f (x) = x^{3} + 1

Frequently Asked Questions (FAQ)

What is the range of f(x)=x^2+16x+8 ?
The range of f(x)=x^2+16x+8 is f(x)>=-56