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Popular >

range of-16t^2+1700

Solution

range

- 16 t^{2} + 1700

Solution

f (t) \leq 1700

+1

Solution steps

Vertex of

- 16 t^{2} + 1700 :

Maximum

(0, 1700)

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 = - 16,

Vertex

(x_{v}, y_{v}) = (0, 1700)

f (t) \leq 1700

Popular Examples

line (5,0),(1,-1)line

(5, 0), (1, - 1)

intercepts of (5x+20)/(x^2+x-12)intercepts

\frac{5 x + 20}{x ^{2} + x - 12}

domain of f(x)=sqrt(1-x^2)+3 domain

f (x) = 1 - x^{2} + 3

inverse of y=\sqrt[3]{x}inverse

extreme f(x)=x^3-2x^2-4x+10 extreme points

f (x) = x^{3} - 2 x^{2} - 4 x + 10

Frequently Asked Questions (FAQ)

What is the range of-16t^2+1700 ?
The range of-16t^2+1700 is f(t)<= 1700