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Popular >

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

Solution

range

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

Solution

f (x) \geq - 2

+1

Interval Notation

[- 2, \infty)

Solution steps

Vertex of

2 x^{2} - 12 x + 16 :

Minimum

(3, - 2)

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 = 2,

Vertex

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

f (x) \geq - 2

Popular Examples

critical 5x^4-2x^2+2x critical points

5 x^{4} - 2 x^{2} + 2 x

domain of f(x)= 2/(\frac{3){x-1}}domain

f (x) = \frac{2}{\frac{3}{x - 1}}

domain of f(x)=sqrt((x+3)/(x-3))domain

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

critical f(x)=(x-6)^3 critical points

f (x) = (x - 6)^{3}

inverse of 4/(s^2-9)inverse

\frac{4}{s ^{2} - 9}

Frequently Asked Questions (FAQ)

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