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Popular >

range of y=x^2+6x+8

Solution

range

y = x^{2} + 6 x + 8

Solution

f (x) \geq - 1

+1

Interval Notation

[- 1, \infty)

Solution steps

Vertex of

x^{2} + 6 x + 8 :

Minimum

(- 3, - 1)

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 = 1,

Vertex

(x_{v}, y_{v}) = (- 3, - 1)

f (x) \geq - 1

Popular Examples

y = lo g_{3} (x)

critical (x+5)/(x^2-12x+35)critical points

\frac{x + 5}{x ^{2} - 12 x + 35}

f (x) = - 3 x^{2}

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

f (x) = x + 5 - 2

line (2,1),(3,2)line

(2, 1), (3, 2)

Frequently Asked Questions (FAQ)

What is the range of y=x^2+6x+8 ?
The range of y=x^2+6x+8 is f(x)>=-1