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Popular >

range of 2x^2-x+4

Solution

range

2 x^{2} - x + 4

Solution

f (x) \geq \frac{31}{8}

+1

Interval Notation

[\frac{31}{8}, \infty)

Solution steps

Vertex of

2 x^{2} - x + 4 :

Minimum

(\frac{1}{4}, \frac{31}{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{31}{8})

f (x) \geq \frac{31}{8}

Popular Examples

domain of sqrt(x)+sqrt(7-x)domain

x + 7 - x

inverse of f(x)=(x^5)/6+7 inverse

f (x) = \frac{x ^{5}}{6} + 7

extreme y=(x-3)^2(x+4)^2 extreme points

y = (x - 3)^{2} (x + 4)^{2}

domain of f(x)=sqrt(\sqrt{x-4)-4}domain

f (x) = x - 4 - 4

slope of 6x+3y=2 slope

6 x + 3 y = 2

Frequently Asked Questions (FAQ)

What is the range of 2x^2-x+4 ?
The range of 2x^2-x+4 is f(x)>= 31/8