In trigonometry, how do I find these inverses?
In trigonometry, would these equations have inverses?
1. x=y^2-3
2. x=abs(y-2)+1
3. x^2+y^2=5
4. x^2-2y^2=9
5. abs(x+y)=4
September 18, 2012
In trigonometry, how do I find these inverses?
In trigonometry, would these equations have inverses?
1. x=y^2-3
2. x=abs(y-2)+1
3. x^2+y^2=5
4. x^2-2y^2=9
5. abs(x+y)=4
6 Comments • Newest first
My question is what do these problems have to do with trigonometry...
[quote=meeop13]So I have to graph x=y^2-3. To graph this, I made it y=+ - sqrt(x+3).
I have to graph the inverse, but I made the original equation into it's inverse. What do I do?
Thanks.[/quote]
Sometimes it's easier to graph the given equation and then use this method to graph the inverse. x=y^2 - 3 is much easier to graph than y = +- sqrt(x+3). Ignoring where the variables are, you can kinda just tell it's simpler. No square roots, so any values you pick for the given equation should come out as whole numbers, as long as you plug in whole numbers.
The thing to note is that when you make up ordered pairs (coordinates) for x = y^2 - 3 is that you're starting by making up y values, not x values. So don't get confused when graphing. I suggest making a table of 4 or 5 y values and quickly solving for all their x values. Once you have a table filled out, make ordered pairs out of your results and graph them.
All that is graphing the given equation. To graph the inverse of that equation... check this out: http://www.purplemath.com/modules/invrsfcn.htm
This should save you a lot of time, compared to using the inverse equation.
(Someone let me know if I'm completely right or wrong at any point... I haven't done inverse equations in a while)
[quote=meeop13]So I have to graph x=y^2-3. To graph this, I made it y=+ - sqrt(x+3).
I have to graph the inverse, but I made the original equation into it's inverse. What do I do?
Thanks.[/quote]
it doesn't have an inverse but! you can draw the graph this way:
http://www.wolframalpha.com/input/?i=y%3DRe%28%2Bsqrt%28x%2B3%29%29+and+y%3DRe%28-sqrt%28x%2B3%29%29++from+-6+to+6
switch x and y, draw, rotate the graph 90 degrees (clock)
So I have to graph x=y^2-3. To graph this, I made it y=+ - sqrt(x+3).
I have to graph the inverse, but I made the original equation into it's inverse. What do I do?
Thanks.
[quote=wrabacon7]Don't you just switch the x and y variables and then solve for y all over again? Sorry if I'm wrong, it's been a while xD[/quote]
this is correct.
just switch the x and y and solve for y, right?