Monday

September 1, 2014

September 1, 2014

Posted by **Marissa** on Tuesday, August 21, 2007 at 11:20pm.

2)Determine the equation of any vertical asymtotes of the graph of f(x)=2x+3/x^2+2x-3

1)

Area of a triangle is 1/2 b*h=A

use the formula and if you have any trouble post your work and I'll see If I can help.

24? thanks for all your help

1) What they are saying is that the area is proptional to the product b*h. If bh decreases by a factor 48/150, the area increases by the same factor. The smaller triangle area is

(48/150)*75 = 24

You could just as easily used the formula for the area of any triangle, (1/2) b h, but they didn't want you to do it that way for some reason. Blame your state's math curriculum committee.

(2) Asymptotes are straight lines that you approach when x becomes very large, whether positive or negative. The way you have written the equation, the 3/x^2 term becomes negligile at large |x|, and the equation becomes y = 4x - 3. That asymptote is not vertical. Is there supposed to be more terms than x^2 in the denominator following 3/ ?. If so, please indicate the denominator with parentheses

#1

"The area A of a triangle varies jointly as the lengths of its base b and height h" means Area = k(b)(h)

so you are given Area=75, b=15, h=10 gets you k=1/2

so now we know the area of a triangle formula:

A=(1/2)(b)(h)

sub in the new values to get A=1/2(8)(6) = 24

#2

vertical asymptotes are caused by the denominator becoming zero

look at the denominator, it factors to

(x+3)(x-1)

so it becomes zero when x=-3 or x=1

so the equations of the verical asymtotes are

x=-3, x=1

I dont know. The top is 2x+3 and the bottom is x^2+2x-3

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