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Popular Functions >

range of f(x)=2x^2-5x+1

Solution

range

f (x) = 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}

Graph

Popular Examples

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parallel y=2x+3,(3,1)

domain of (sqrt(x-3))^2

Frequently Asked Questions (FAQ)

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