# Thread: Help with functons and line and circle formulas

1. ## Help with functons and line and circle formulas

Few questions from my homework.

Find the equation of a circle with diameter endpoints (8,10) and (-2,14). I know I need to find the distance and midpoint but am having trouble with putting togther the exact formula.

2. Find the equation of a line in slope intercept form passing through (-2,-5) and (3,5)

And then find equation of a line that is parallel to the line above passing through (-3,4)

Functions:
How would you graph f(x)= sqaure root of x+2 - 3 Note: the 3 is not under the square root.

Explain what is being done to f(x) in this equation y=-2f(x+3)-1

How would you show that f(x)=8x^3 -3x is odd.

Thanks to anyone who can help me out.

2. ## The last one...

The def of an odd function is $f(-x)=-f(x)$

$f(x)=8x^3-3x$

$f(-x)=8(-x)^3-3(-x) = -8x^3+3x$

if we factor out a minus sign

$f(-x)=8(-x)^3-3(-x) = -8x^3+3x=-\underbrace{(8x^3-3x)}_{\mbox{This is f(x)}}=-f(x)$

so by the above definition it is an odd function.

3. Thanks for the help with that one, any idea with the others?

4. ## #2

we use this formula to find the slope

$m=\frac{y_2-y_1}{x_2-x_1}=\frac{-5-5}{-2-3}=\frac{-10}{-5}=2$

now we can use the slope intercept from to find the equation of the line.

$y=mx+b$ we know m from our above work so

$y=2x+b$ now using either point (3,5)

$5=2(3)+b \iff -1 =b$ so the equation is

$y=2x-1$

For part b use the same slope and procedure with the new point (-3,4)