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Popular >

range of f(x)=-16x^2+48x+160

Solution

range

f (x) = - 16 x^{2} + 48 x + 160

Solution

f (x) \leq 196

+1

Interval Notation

(- \infty, 196]

Solution steps

Vertex of

- 16 x^{2} + 48 x + 160 :

Maximum

(\frac{3}{2}, 196)

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 = - 16,

Vertex

(x_{v}, y_{v}) = (\frac{3}{2}, 196)

f (x) \leq 196

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(x - 1)^{2}

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\frac{4 x + 9}{3 x - 6}

inverse of t^2 inverse

t^{2}

domain of f(x)=5*2^x domain

f (x) = 5 \cdot 2^{x}

domain of sqrt((4-x^2)/(x+1))domain

\frac{4 - x ^{2}}{x + 1}

Frequently Asked Questions (FAQ)

What is the range of f(x)=-16x^2+48x+160 ?
The range of f(x)=-16x^2+48x+160 is f(x)<= 196