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Popular >

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}

Popular Examples

critical f(x)=t^4-24t^3+154t^2 critical points

f (x) = t^{4} - 24 t^{3} + 154 t^{2}

inverse of 7/(x^2+1)inverse

\frac{7}{x ^{2} + 1}

intercepts of f(x)=(4x+9)/(3x-6)intercepts

f (x) = \frac{4 x + 9}{3 x - 6}

intercepts of y=x+7 intercepts

y = x + 7

inverse of f(x)=e^{(sqrt(x))/3}inverse

f (x) = e^{\frac{x}{3}}

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