This week we learned about 3 conics. We learned about circles, ellipses, and hyperbola's. Last year I nearly failed the final exam b/c of conics, but now I think they are pretty simple. I am going to explain how to put a circle in standard form. To do this I will use a form of completing the square.
Ex. x^2+y^2-6x+4y-12=0
1.)Put your like terms together and leave space:
x^2-6x+__+y^2+4y+__=0
2.)Take your linear coefficients and divide by two then square. Add results to each side in respective blanks:
x^2-6x+9+y^2+4y+4=12+9+4
3.)Factor and Simplify:
(x-3)^2+(y+2)^2=25
And that is standard form of a circle.
Standard Form is: (x-h)^2+(y-k)^2=r^2; where (h,k) is the center and r is the radius.
***If you want to find the center or radius, put equation in standard form and take respective elements.****
I understood all of the conic stuff it's just remembering formulas is what will be hard. I looked through my chapter tests and one thing I'm still not sure on how to do is find inverses of functions. Any help is appreciated. Thanks.
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Inverse of a Function
ReplyDeletesteps:
1. Replace f(x) with y
2. Reverse the roles of x and y
3. Solve for y in terms of x
4. Replace y with f-1(x)
Example 1 - f(x) = 2x + 3
write the function as an equation:
y = 2x + 3
solve for x: x = (y - 3)/2
now write f-1(y) as follows:
f -1(y) = (y - 3)/2 or f -1(x) = (x - 3)/2
Check:
f(f -1(x))=2(f -1(x)) + 3
=2((x-3)/2)+3 =(x-3)+3 =x
f -1(f(x))=f -1(2x+3)
=((2x+3)-3)/2 =2x/2 =x
Ill post my notes because i think they should be neat and simple enough for you to pick up on inverses
ReplyDelete* to find an inverse:
switch X & Y
solve for Y
before you switch the X and Y
you'll have to check to see if the equation passes the horizontal line test
the horizontal line test consists of sketching a graph of the equation then drawing a horizontal line anywhere on the graph if the sketch of the graphed equation only touches the line once then it passes the horizontal line test and you can proceed to switching the X and the Y
however if the sketch of the graphed equation touches the line more than once then it does not pass the horizontal line test.
If an equation does not pass the horizontal line test then you dont even have to attempt to solve it all you have to put is NO INVERSE and you're done.
Finally there is the step of proving youre answer is an inverse
for this all you have to do is remember two formulas
#1 f(f^-1(X)) = X
#2 f^-1(f(X)) = X
now for youre example
EX: y=5x-2
• Sketch a graph for the horizontal line test. * it passes
• Switch the X and Y * x=5y-2
• Solve for Y * x-2=5y therefore y= x+2/5
Next you’ll have to prove
Steps for proving an inverse
• Plug the inverse in *you’ll plug it into the first formula
* this formula is f(f-1(x)) = X
• Plug the inverse in * you’ll plug it into the second formula
* this formula is f-1(f(x)) = X
Ex: *y=5(x+2/5)-2 = x+2-2 = x
*y=(5x-2)/5+2 = x-2+2 = x
If each of the answers to the formulas are both = X then you completed the problem correctly
for inverses you first you check to see if it passed the horizontal line test, if it does then you will switch X and Y and you will solve for Y.
ReplyDelete-Y=square root of X+1
-X=square root of Y+1
-square booth and get
-X^2=Y+1
-X^2-1=Y
Y=X^2
then to prove you would just plug it in