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Popular >

range of 4x^2-5x+7

Solution

range

4 x^{2} - 5 x + 7

Solution

f (x) \geq \frac{87}{16}

+1

Interval Notation

[\frac{87}{16}, \infty)

Solution steps

Vertex of

4 x^{2} - 5 x + 7 :

Minimum

(\frac{5}{8}, \frac{87}{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{5}{8}, \frac{87}{16})

f (x) \geq \frac{87}{16}

Popular Examples

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

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

asymptotes of 3/(x+2)asymptotes

\frac{3}{x + 2}

domain of f(x)=sqrt(\sqrt{x-2)+5}domain

f (x) = x - 2 + 5

slope ofintercept-4x+y=2 slope intercept

- 4 x + y = 2

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

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

Frequently Asked Questions (FAQ)

What is the range of 4x^2-5x+7 ?
The range of 4x^2-5x+7 is f(x)>= 87/16