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Finding the inverse of an equations requires a few steps. I will use the orginal function f(x)=2x2+2x1

 

1. Change f(x) to y.

 

This means that f(x)=2x2+2x1 changes to y=2x2+2x1. That is all that this step entails.

 

2Replace all instances of y with x and all instances of y with x

 

This step is relatively simple, too. 

 

y=2x2+2x1 changes to x=2y2+2y1. Now, this step is done.

 

3. Solve for y

 

Solving for y means to isolate it. Since the quadratic formula is only a valid option when solving for the root of the equation, we cannot use that method with multiple variables. It appears as if completing the square is the only option.

 

2y2+2y1=xWe need to get the "c" term on the opposite side of the equation for completing the square. Therefore, add 1 to both sides.
2y2+2y=x+1Since the a-term must be one in order for completing the square to work, we must divide the entire equation by 2.
y2+y=x+12This is the trickiest bit. We need to make the lefthand side a perfect square. To do this, add (b2)2 where b is the coefficient of the linear term. You must add it to both sides because whatever you do to one side, you must do to the other.
y2+y+(b2)2=x+12+(b2)2Replace b with the coefficient of the linear term, 1.
y2+y+(12)2=x+12+(12)2Simplify both sides.
y2+y+14=x+12+14The left hand side is now a perfect-square, so transform it into one. 
(y+12)2=x+12+14Before taking the square root of both sides, we should add the fractions together.
x+12+14=2x+24+14=2x+34Reinsert this back into the equation.
(y+12)2=2x+34Take the square root of both sides.
y+12=±2x+34Distribute the square root to both the numerator and denominator. Remember that taking the square root results in the positive and negative answer.
y+12=±2x+32Subtract 1/2 from both sides.
y=±2x+312Break them up into separate solutions.
y=2x+312y=2x+3+12

 

We can convert this into function notation, if you want.
f1(x)=±2x+312 

 

I will get to graphing later.

 

The next function is f(x)=4.9(t+3)2+45.8 

 

Let's do the same steps again. Change f(x) to y and flip flop all "t's" and "y's."

 

t=4.9(y+3)2+45.8

 

Last time, I solved the equation, but I won't do that with this function because it appears as if you did it correctly. 

 

Ok, inputting it into Desmos is easier than you might think. 

 

Equation 1: y=2x2+2x1

 

Equation 1 Inverse: x=2y2+2y1

 

This is the inverse because every x value is now the y-value. We can do the same for equation 2/

 

Equation 2: y=4.9(t+3)2+45.8

 

Equation 2 Inverse: t=4.9(y+3)2+45.8

 

Now, just put in both the "Equation 1 Inverse" and "Equation 2 Inverse" into Desmos, and Desmos figure out the rest. Click here to see the graph is Desmos. Spoiler Alert: both equations fail the vertical line test.

Sep 16, 2017
 #1
avatar+590 
+1
Sep 16, 2017
Sep 15, 2017

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