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Popular >

range of 2x^2+x-14

Solution

range

2 x^{2} + x - 14

Solution

f (x) \geq - \frac{113}{8}

+1

Interval Notation

[- \frac{113}{8}, \infty)

Solution steps

Vertex of

2 x^{2} + x - 14 :

Minimum

(- \frac{1}{4}, - \frac{113}{8})

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}) = (- \frac{1}{4}, - \frac{113}{8})

f (x) \geq - \frac{113}{8}

Popular Examples

critical f(x)=4x^{1/3}-x^{4/3}critical points

f (x) = 4 x^{\frac{1}{3}} - x^{\frac{4}{3}}

inverse of f(x)=2x^5 inverse

f (x) = 2 x^{5}

inverse of f(x)=sqrt(x+3)-9 inverse

f (x) = x + 3 - 9

asymptotes of f(x)=(x-3)/(x+3)asymptotes

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

inflection (X^4)/((1+X)^3)inflection points

\frac{X ^{4}}{( 1 + X ) ^{3}}

Frequently Asked Questions (FAQ)

What is the range of 2x^2+x-14 ?
The range of 2x^2+x-14 is f(x)>=-113/8