+2446 Finding the inverse of an equations requires a few steps. I will use the orginal function \(f(x)=2x^2+2x-1\)
1. Change \(f(x)\) to \(y\).
This means that \(f(x)=2x^2+2x-1\) changes to \(y=2x^2+2x-1\). That is all that this step entails.
2. Replace all instances of y with x and all instances of y with x
This step is relatively simple, too.
\(y=2x^2+2x-1\) changes to \(x=2y^2+2y-1\). 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.
| \(2y^2+2y-1=x\) | We need to get the "c" term on the opposite side of the equation for completing the square. Therefore, add 1 to both sides. | ||
| \(2y^2+2y=x+1\) | Since the a-term must be one in order for completing the square to work, we must divide the entire equation by 2. | ||
| \(y^2+y=\frac{x+1}{2}\) | This is the trickiest bit. We need to make the lefthand side a perfect square. To do this, add \(\left(\frac{b}{2}\right)^2\) 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. | ||
| \(y^2+y+\left(\frac{b}{2}\right)^2=\frac{x+1}{2}+\left(\frac{b}{2}\right)^2 \) | Replace b with the coefficient of the linear term, 1. | ||
| \(y^2+y+\left(\frac{1}{2}\right)^2=\frac{x+1}{2}+\left(\frac{1}{2}\right)^2 \) | Simplify both sides. | ||
| \(y^2+y+\frac{1}{4}=\frac{x+1}{2}+\frac{1}{4} \) | The left hand side is now a perfect-square, so transform it into one. | ||
| \(\left(y+\frac{1}{2}\right)^2=\frac{x+1}{2}+\frac{1}{4}\) | Before taking the square root of both sides, we should add the fractions together. | ||
| \(\frac{x+1}{2}+\frac{1}{4}=\frac{2x+2}{4}+\frac{1}{4}=\frac{2x+3}{4}\) | Reinsert this back into the equation. | ||
| \(\left(y+\frac{1}{2}\right)^2=\frac{2x+3}{4}\) | Take the square root of both sides. | ||
| \(y+\frac{1}{2}=\pm\sqrt{\frac{2x+3}{4}}\) | Distribute the square root to both the numerator and denominator. Remember that taking the square root results in the positive and negative answer. | ||
| \(y+\frac{1}{2}=\pm\frac{\sqrt{2x+3}}{2}\) | Subtract 1/2 from both sides. | ||
| \(y=\pm\frac{\sqrt{2x+3}-1}{2}\) | Break them up into separate solutions. | ||
| We can convert this into function notation, if you want. | ||
| \(f^{-1}(x)=\pm\frac{\sqrt{2x+3}-1}{2}\) |
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: \(y=2x^2+2x-1\)
Equation 1 Inverse: \(x=2y^2+2y-1\)
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.
+2446 In order to convert an equation in slope-intercept form (\(y=mx+b\)) into standard form, you must understand a few rules. I will list them for you:
1. \(Ax+Bx=C\) is the form of the linear function
2. \(A,B,C\in\mathbb{Z}\) (A, B, and C must be integers)
3. \(A\geq0\)
4. A,B, and C must be co-prime
Knowing these rules, let's now convert \(y=-\frac{2}{5}x-5\) into standard form:
| \(y=-\frac{2}{5}x-5\) | Because standard form does not allow there to be fractions, let's eliminate the current fraction by multiplying all sides by 5. |
| \(5y=-2x-25\) | Add 2x to both sides. |
| \(2x+5y=-25\) | We have converted the equation from slope-intercept form while meeting all the conditions. |
+2446 This process is more or less the same when you have 2 arbitrary points (such as (4,6) and (-9,1)). This time, however, we must take into account that there are variables involved. Let's remind you of slope-intercept form of a line.
\(y=mx+b\)
m = slope of the line
b = y-intercept
In this particular case, we know the x- and y-intercepts because those points are given in the original problem. We know that the y-intercept is located at \((0,b)\). Since b is the y-intercept, fill that in! That's the easy bit, I think you'd agree.
\(y=mx+b\)
We know that the x-intercept is at the point when y=0, so plug that in:
| \(0=mx+b\) | Now, solve for x by subtracting b on both sides. |
| \(-b=mx\) | Divide by m on both sides. |
| \(x=\frac{-b}{m}\) | |
We have determined, with the above algebraic work that when \(y=0,\hspace{1mm}x=\frac{-b}{m}\), which means that the x-intercept is located at \(\left(\frac{-b}{m},0\right)\). However, we also know that the x-intercept is located at \((a,0)\), which means that \(a=\frac{-b}{m}\):
| \(a=\frac{-b}{m}\) | Now, we must solve for m because that is the slope of this linear equation after all. |
| \(ma=-b\) | Divide by a on both sides. |
| \(m=\frac{-b}{a}\) | |
We now know the value for b and for m, so fill it in to get the equation.
\(y=\frac{-b}{a}x+b\)
.+2446 Your question appears to want one to evaluate the expression \(\sqrt{(108b)^4}\). Let me try to make it easier so that the calculator is not necessary:
| \(\sqrt{(108b)^4}\) | Of course, the square root can also be represented as the power to 1/2. |
| \(\left((108b)^4\right)^\frac{1}{2}\) | Using the power rule, we know that \(\left(a^b\right)^c=a^{b*c}\). Let's apply that. |
| \(\left((108b)^4\right)^\frac{1}{2}=(108b)^{4*\frac{1}{2}}=(108b)^2\) | Now, distribute the exponent. |
| \((108b)^2=108^2*b^2\) | Now, let's simplify 108^2 by doing this. |
| \(108^2=(100+8)(100+8)=10000+800+800+64=11664\) | This, to me, is the easiest way to calculate the square of a number without a calculator. What do you think? |
| \(11664b^2\) | This is your final answer. |
