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Popular Functions >

range of f(x)=4x^2+5x-1

Solution

range

f (x) = 4 x^{2} + 5 x - 1

Solution

f (x) \geq - \frac{41}{16}

+1

Interval Notation

[- \frac{41}{16}, \infty)

Solution steps

Vertex of

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

Minimum

(- \frac{5}{8}, - \frac{41}{16})

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 = 4,

Vertex

(x_{v}, y_{v}) = (- \frac{5}{8}, - \frac{41}{16})

f (x) \geq - \frac{41}{16}

Graph

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Frequently Asked Questions (FAQ)

What is the range of f(x)=4x^2+5x-1 ?
The range of f(x)=4x^2+5x-1 is f(x)>=-41/16