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Popular >

range of 2x^2+5x+10

Solution

range

2 x^{2} + 5 x + 10

Solution

f (x) \geq \frac{55}{8}

+1

Interval Notation

[\frac{55}{8}, \infty)

Solution steps

Vertex of

2 x^{2} + 5 x + 10 :

Minimum

(- \frac{5}{4}, \frac{55}{8})

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 = 2,

Vertex

(x_{v}, y_{v}) = (- \frac{5}{4}, \frac{55}{8})

f (x) \geq \frac{55}{8}

Popular Examples

symmetry y=x^2+4x+2 symmetry

y = x^{2} + 4 x + 2

asymptotes of f(x)=(x-8)/(x^2-64)asymptotes

f (x) = \frac{x - 8}{x ^{2} - 64}

intercepts of y=1-x^2 intercepts

y = 1 - x^{2}

simplify (2.1)(-6.7)simplify

(2.1) (- 6.7)

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

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

Frequently Asked Questions (FAQ)

What is the range of 2x^2+5x+10 ?
The range of 2x^2+5x+10 is f(x)>= 55/8