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Popular >

range of g(x)=-(x+5)^2+2

Solution

range

g (x) = - (x + 5)^{2} + 2

Solution

f (x) \leq 2

+1

Solution steps

Vertex of

- (x + 5)^{2} + 2 :

Maximum

(- 5, 2)

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, 2)

f (x) \leq 2

Popular Examples

inverse of y= 1/(x+2)inverse

y = \frac{1}{x + 2}

extreme f(x)=2x^3-3x^2-12x+4 extreme points

f (x) = 2 x^{3} - 3 x^{2} - 12 x + 4

inverse of f(x)=(2x-3)^2+1 inverse

f (x) = (2 x - 3)^{2} + 1

midpoint (3,6),(5.5,5)midpoint

(3, 6), (5.5, 5)

inflection y=(x^3)/3-2x^2-12x inflection points

y = \frac{x ^{3}}{3} - 2 x^{2} - 12 x

Frequently Asked Questions (FAQ)

What is the range of g(x)=-(x+5)^2+2 ?
The range of g(x)=-(x+5)^2+2 is f(x)<= 2