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Popular >

range of f(x)=16-(12x+8)^2

Solution

range

f (x) = 16 - (12 x + 8)^{2}

Solution

f (x) \leq 16

+1

Interval Notation

(- \infty, 16]

Solution steps

Vertex of

16 - (12 x + 8)^{2} :

Maximum

(- \frac{2}{3}, 16)

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 = - 144,

Vertex

(x_{v}, y_{v}) = (- \frac{2}{3}, 16)

f (x) \leq 16

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

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f (x) = x + x^{2} + \frac{1}{x ^{2}}

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7 - 3 x

Frequently Asked Questions (FAQ)

What is the range of f(x)=16-(12x+8)^2 ?
The range of f(x)=16-(12x+8)^2 is f(x)<= 16