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Popular >

range of f(x)=x^2+8x+15

Solution

range

f (x) = x^{2} + 8 x + 15

Solution

f (x) \geq - 1

+1

Interval Notation

[- 1, \infty)

Solution steps

Vertex of

x^{2} + 8 x + 15 :

Minimum

(- 4, - 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 = 1,

Vertex

(x_{v}, y_{v}) = (- 4, - 1)

f (x) \geq - 1

Popular Examples

amplitude of-cos(2(θ-pi/4))amplitude

- cos (2 (θ - \frac{π}{4}))

critical f(x)=3x^5-20x^3 critical points

f (x) = 3 x^{5} - 20 x^{3}

domain of f(x)=(8-x)/(x^2-7x)domain

f (x) = \frac{8 - x}{x ^{2} - 7 x}

inverse of f(x)=7x inverse

f (x) = 7 x

(12), (14)

Frequently Asked Questions (FAQ)

What is the range of f(x)=x^2+8x+15 ?
The range of f(x)=x^2+8x+15 is f(x)>=-1