The steps are not too long-winded luckily. Unfortunately for Cphill, the 3 that precedes the argument is missing from Cphill's answer.
1) As Cphill mentioned, if \(f(x)\) is the original function, then \(f(x-6)\) translates the function 6 units rightward. The instructions specificially request for the answer with an equation of \(g(x)\) in terms of \(f(x)\), \(g(x)=f(x-6)\)
2) In the previous answer, we determined that \(g(x)=f(x-6)\), so we can use this information to solve for the current problem:
\(g(x)=f(x-6)\\ f(x)=3(x-5)^2+9\) | \(f(x)=3(x-5)^2+9\) |
\(f(x-6)=3[(x-6)-5]^2+9\) | |
\(f(x-6)=3(x-11)^2+9\) | |
\(g(x)=f(x-6)=3(x-11)^2+9\\ g(x)=3(x-11)^2+9\) |
3) If \(g(x)\) is translated 7 units downward to obtain \(j(x)\), then \(g(x)-7=j(x)\). Of course, the question asks for the equation to be in terms of f(x):
\(j(x)=g(x)-7\) | Now, substitute g(x)=f(x-6). |
\(j(x)=f(x-6)-7\) | |
4) Find j(x) in terms of x:
\(j(x)=g(x)-7\) | We have determined previously that \(g(x)=3(x-11)^2+9\), so substitute that in. |
\(j(x)=[3(x-11)^2+9]-7\) | Now, simplify. There is not much to do. |
\(j(x)=3(x-11)^2+2\) | |
The second one was definitely much harder for me to come up with! Here is a picture that may be useful to reference as I solve:
You will probably notice that I added a circle that inscribes the triangle. At its bost basic level, the circumcenter is a point that is equidistant from all the vertices of the triangle. The circumcenter is also the center of a triangle's circumcircle, which I have illustrated with the diagram above. Of course, we know that the sum of \(\angle AHC\) and \(\angle AOC\) is 240 by the given info. Notice that \(m\angle AOC=m\angle AHC\) because they both intercept the same arc. Therefore, we can solve for the angle of the circumcenter.
\(m\angle AOC+m\angle AHC=240\) | Since we already established that \(m\angle AOC=m\angle AHC\), we can solve for the the central angle of the circle. |
\(m\angle AOC+m\angle AOC=240\) | |
\(2m\angle AOC=240\) | |
\(m\angle AOC=120^\circ\) | There is no need to solve for the angle measure of the orthocenter; it is simply a waster of time. |
Notice that \(\angle ABC\) is an inscribed angle, which means that its measure is equal to half the measure of the central angle.
\(\frac{1}{2}m\angle AOC=m\angle ABC\) | Use substitution here since we already know the measure of one of the angles. |
\(\frac{1}{2}*120=m\angle ABC\) | |
\(m\angle ABC=60^\circ\) |
I have been thinking about these two problems for some time, and I think I have finally cracked them! I'll now impart my knowledge to you!
1)
\(\triangle ABC\text{ is an isosceles right triangle}\\ m\angle A=90^{\circ}\\ O\text{ is the circumcenter}\) | Given information |
\(m\angle A+m\angle B+m\angle C=180^{\circ}\) | Triangle sum theorem (the sum of the interior angles of a triangle equals 180 degrees) |
\(90^{\circ}+m\angle B+m\angle C=180^{\circ}\) | Substitution property of equality |
\(m\angle B+m\angle C=90^\circ\) | Subtraction property of equality |
\(m\angle B=m\angle C\) | Isosceles Triangle Theorem (The angles opposite congruent sides in an isosceles triangle are congruent) |
\(m\angle B+m\angle B=90^\circ\) | Substitution property of equality |
\(2m\angle B=90^\circ\) | Simplify |
\(m\angle B=45^\circ\) | Division property of equality |
\(\overline{AO}\cong\overline{BO}\) | Property of Circumcenter (The circumcenter is equidistant from the vertices of the triangle) |
\(m\angle A=m\angle B\) | |
Well, this is taking some time...
Anyway, \(\triangle AOB\) is another isosceles triangle. \(m\angle OAB=m\angle B=45^\circ\), so the remaining angle, \(\angle BOA=90^\circ\)
.By the given information, we know the following:
\(t(x)=\sqrt{3x+1}\\ f(x)=5-t(x)\)
In order to find \(t(f(5))\), we must first evaluate f(x) when x=5:
\(f(x)=5-t(x)\) | Since t(x) appears in the definition of f(x), plug in t(x) into the f(x) function. |
\(f(x)=5-\sqrt{3x+1}\) | Now, evaluate f(x) when x=5. |
\(f(5)=5-\sqrt{3*5+1}\) | Notice how every instance of x has been replaced wth a 5. Now, it is a matter of simplifying. |
\(f(5)=5-\sqrt{16}\) | |
\(f(5)=5-4\) | |
\(f(5)=1\) | |
Now we know that \(t(f(5)=t(1)\) since we just determined that \(f(5)=1\).
\(t(x)=\sqrt{3x+1}\) | Of course, we want to evaluate when x=1, so replace every instance of x with a 1. |
\(t(1)=\sqrt{3*1+1}\) | Now, simplify. |
\(t(1)=\sqrt{3+1}\) | |
\(t(1)=\sqrt{4}=2\) | |
Therefore, \(t(f(5))=2\)
.Understandably, square roots cause a lot of confusion. This is one prime example of this occurring. To be clear, by definition, \(i=\sqrt{-1}\). When you calculate the square root of a number, it is implied that the output is the principal root or the positive result.
For example,
\(\sqrt{4}=2\\ \sqrt{9}=3\)
However, if you place a negative sign before the square root, then it is implied that the output should be negative.
\(-\sqrt{4}=-2\\ -\sqrt{9}=-3\)
If a plus-minus sign precedes a square root, then both answers are desired:
\(\pm\sqrt{4}=\pm2\\ \pm\sqrt{9}=\pm3\)
This confuses many people, and it leads to misconceptions. Since \(\sqrt{-1}\) implies the principal root, the only answer is \(i\).
Examine the false proof that your "guy" provided, which seems to prove that i equals plus or minus 1 to the untrained eye.
\(i^2\\ i^\frac{4}{2}\\ \left(i^4\right)^{0.5}\\ i^{(2+2)^{0.5}}\\ \textcolor{red}{\left(i^2*i^2\right)^{0.5}}\)
The step highlighted in red in an invalid step. It ignores the part about exponent towers. In line 4, you have to evaluate the innermost exponent first to abide by order of operations. In line 5, there are now parentheses around the \(i^2*i^2\), which changes the overall value of the expression. If this is unclear for you, maybe I will try a simpler example.
Consider the following two expressions \(2^{3^2}\text{ and }\left(2^3\right)^2\)? The expressions here have a subtle difference, but the output is different. You compute these differently, too. This is another area where misconceptions are prevalent.
\(2^{\textcolor{red}{3^2}}=2^9=512\\ \textcolor{red}{\left(2^3\right)}^2=8^2=64\)
Notice that in the second expression that the parentheses take precedence over any other operation, so you must do that first. Now, let's relate that to the false proof.
\(i^{(2+2)^{0.5}}=\textcolor{red}{i^{4^{0.5}}}=i^2=-1\\ (i^2*i^2)^{0.5}=(i^{2+2})^{0.5}=\textcolor{red}{\left(i^4\right)^{0.5}}\)
Do you notice the discrepancy now? You can now see that the values are different. Since I have cleared up that confusion, I will now end the debate about which complex number the product simplifies to. I'll jump right to the moment of the confusion, which occurs at \(18+3i-3i^2\)
\(18+3i-3i^2\) | By definition, \(i=\sqrt{-1}\), so squaring both sides would mean that \(i^2=-1\). |
\(18+3i-3*-1\) | Now, simplify as much as possible. |
\(18+3i+3\) | Combine the like terms. |
\(21+3i\) | This is completely simplified, and nothing more can be done here. |
Well, trial and error appeared to be my friend here. If I was not aware that the original expression could be simplified further, I probably would have made the same conclusion as you did, Melody. It is considered improper to have radicals or fractional exponents in the denominator, so I knew that there was some way to simplify this. My method is really only relevant to this particular problem. You will see why.
Now, let's consider that extra bit on the end, the \(9^\frac{1}{3}=\left(3^2\right)^\frac{1}{3}=3^\frac{2}{3}=3^{\left(\frac{1}{3}\right)^2}\). This means that the denominator can be written like \(1+3^\frac{1}{3}+3^{\left(\frac{1}{3}\right)^2}\), and if I let \(x=3^\frac{1}{3}\), then I can represent the denominator like \(x^2+x+1\). I then made a stunning realization. This required some extraordinary observational skills.
\(x^3-y^3=(x-y)(x^2+xy+y^2)\) | I am sure that you are familiar with this. It's the factorization of a difference of cubes. Let's set y equal to 1 and substitute. |
\(x^3-1=(x-1)(x^2+x+1)\) | Wait a second! Look at what one of the factors is. It's \(x^2+x+1\), which is also the expression of the denominator. This tells me that if I multiply the trinomial in the denominator by x-1, then I will be left with two terms. This is perfect! I have now found a way to manipulate the denominator into two terms. Let's keep going. Since \(x=3^\frac{1}{3}\), let's just substitute it in. |
\(3^{\left(\frac{1}{3}\right)^3}-1=\left(3^\frac{1}{3}-1\right)\left(3^{\left(\frac{1}{3}\right)^2}+3^\frac{1}{3}+1\right)\) | Let's complete the simplification here. The process is quite straightforward when you are comfortable with the law of indices. |
\(3^{\left(\frac{1}{3}\right)^3}-1\\ 3^{\frac{1}{3}*3}-1\\ 3^1-1\\ 3-1\\ 2\) | Of course, notice what I multiplied the denominator by in the original problem. It is \(3^\frac{1}{3}-1\). |
So, no, the conjugate is not "magical." The only issue with this thought process is that this will not help you on most types of problems with multi-term denominators. This is just one particular case, and I happened to crack it.