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Popular >

range of f(x)=4x^2+x+4

Solution

range

f (x) = 4 x^{2} + x + 4

Solution

f (x) \geq \frac{63}{16}

+1

Interval Notation

[\frac{63}{16}, \infty)

Solution steps

Vertex of

4 x^{2} + x + 4 :

Minimum

(- \frac{1}{8}, \frac{63}{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 = 4,

Vertex

(x_{v}, y_{v}) = (- \frac{1}{8}, \frac{63}{16})

f (x) \geq \frac{63}{16}

Popular Examples

asymptotes of 1/(x-2)+1 asymptotes

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

inverse of f(x)= 6/(sqrt(8-x))inverse

f (x) = \frac{6}{8 - x}

intercepts of f(x)=2x^2+8x-10 intercepts

f (x) = 2 x^{2} + 8 x - 10

intercepts of y=3x-9 intercepts

y = 3 x - 9

domain of f(x)=(sqrt(x+5))/(x-6)domain

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

Frequently Asked Questions (FAQ)

What is the range of f(x)=4x^2+x+4 ?
The range of f(x)=4x^2+x+4 is f(x)>= 63/16