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Popular >

range of x^2-10x+16

Solution

range

x^{2} - 10 x + 16

Solution

f (x) \geq - 9

+1

Interval Notation

[- 9, \infty)

Solution steps

Vertex of

x^{2} - 10 x + 16 :

Minimum

(5, - 9)

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}) = (5, - 9)

f (x) \geq - 9

Popular Examples

parallel y=-1/4 x+5 parallel

y = - \frac{1}{4} x + 5

inverse of 9x-5 inverse

9 x - 5

domain of f(x)=(x^2-x-2)/(x-2)domain

f (x) = \frac{x ^{2} - x - 2}{x - 2}

critical f(x)=2x^2-6x^4 critical points

f (x) = 2 x^{2} - 6 x^{4}

asymptotes of f(x)= 1/(x(x+3))asymptotes

f (x) = \frac{1}{x ( x + 3 )}

Frequently Asked Questions (FAQ)

What is the range of x^2-10x+16 ?
The range of x^2-10x+16 is f(x)>=-9