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Popular Functions >

range of f(x)=-5x^2+2x+8

Solution

range

f (x) = - 5 x^{2} + 2 x + 8

Solution

f (x) \leq \frac{41}{5}

+1

Interval Notation

(- \infty, \frac{41}{5}]

Solution steps

Vertex of

- 5 x^{2} + 2 x + 8 :

Maximum

(\frac{1}{5}, \frac{41}{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 = - 5,

Vertex

(x_{v}, y_{v}) = (\frac{1}{5}, \frac{41}{5})

f (x) \leq \frac{41}{5}

Graph

Popular Examples

domain of sqrt(4-6a)

inverse of f(x)=(-2x-5)/(6x-7)

critical f(x)=e^{-2.5x^2}

range of f(x)= 1/(sqrt(x^2+1))

Frequently Asked Questions (FAQ)

What is the range of f(x)=-5x^2+2x+8 ?
The range of f(x)=-5x^2+2x+8 is f(x)<= 41/5