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Popular >

range of 2x^2-5x+1

Solution

range

2 x^{2} - 5 x + 1

Solution

f (x) \geq - \frac{17}{8}

+1

Interval Notation

[- \frac{17}{8}, \infty)

Solution steps

Vertex of

2 x^{2} - 5 x + 1 :

Minimum

(\frac{5}{4}, - \frac{17}{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{17}{8})

f (x) \geq - \frac{17}{8}

Popular Examples

inverse of y=(1-e^{-x})/(1+e^{-x)}inverse

y = \frac{1 - e ^{- x}}{1 + e ^{- x}}

intercepts of (-3x-9)/(5x+15)intercepts

\frac{- 3 x - 9}{5 x + 15}

domain of g(x)= x/5 domain

g (x) = \frac{x}{5}

inverse of f(x)=-4/5 x-12 inverse

f (x) = - \frac{4}{5} x - 12

intercepts of f(x)=(x^2-x-12)/(2x-8)intercepts

f (x) = \frac{x ^{2} - x - 12}{2 x - 8}

Frequently Asked Questions (FAQ)

What is the range of 2x^2-5x+1 ?
The range of 2x^2-5x+1 is f(x)>=-17/8