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Popular >

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}

Popular Examples

inverse of f(x)=5-2/3 x inverse

f (x) = 5 - \frac{2}{3} x

extreme f(x)=9x^2-2x^3 extreme points

f (x) = 9 x^{2} - 2 x^{3}

critical \sqrt[3]{x}(x+4)critical points

inverse of f(x)=e^{arctan(x)}inverse

f (x) = e^{a r c t a n (x)}

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

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

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