hxx2x-43-beginarrayltext-a-identify-the-x-text-intercepts-of-the-graph-of-h-text-b-what-are-the-x-te

Question

$h(x)=(x+2)(x-4)^{3}$
$$
\begin{array}{l}{	ext { (a) Identify the } x 	ext { -intercepts of the graph of } h .} \ {	ext { (b) What are the } x 	ext { -intercepts of the graph of }} \ {y=h(x-2) ?}\end{array}
$$

Ankit Gupta · Accepted Answer

were given a polynomial function. That is it. Your fix is equal Toe X plus two multiplied with X minus four whole cubed now, First of all, we need to find the X intercept for the given function, so they will be at minus two and four repeated trays at four. Since the multiplicity is three off the real solution four. Now we need to find the function first, advise equal do etch Off X minus two. So then they given function will change Toe X minus two plus two multiplied with X minus two minus four. Whole cube, which will become X since minus two plus two will get cancelled, multiplied with X minus six Whole cube. So the new X intercept will now be at zero on six.

Karl Schaefer · Answer

and this problem, we&#x27;re given the functions h of X equals X cubed over four, and K evac equals four x of 1/3 and part A were asked so that they&#x27;re in verses of each other. So what this means is we need to show that h of K of X is equal to X, and we need to show that K of H of X is equal acts. So this is what it means to be in verse is that if you plug in one function to the other than you just get X back. So let&#x27;s let&#x27;s just show it. Let&#x27;s show this 1st 1 h of k of X is you will h of for act two the 1/3 just plugging in K of X. Now let&#x27;s plug this into our formula for H. We get four acts to the 1/3 R cubed over four, so the cubing and the cube root cancel each other. So this is for acts over four, which is just equal act, which is what we wanted. Now the second piece okay of H of X is equal to K o X cubed over for now, we&#x27;re gonna plug this into our expression for K of X. So this is four times X cubed over four, the 1/3 power for cancels with the for the denominator. We&#x27;re left with X cube to the 1/3 and again by exponents, rules the third and the 1/3 camps alive. And we&#x27;re just left with X. So this tells us that the functions are in Versace of each other, and part of you were asked to graph these functions. So let&#x27;s oops, bring up drawing Nice graph here. So there&#x27;s a couple ways that you can approach graphing these functions. You know, you&#x27;re using results of previous section just plugging into a calculator. Something like that. Let&#x27;s remind ourselves what these functions are. H of X equals X cubed over four and K of X is equal to four X to the 1/3. So we&#x27;re going to go ahead and graph finarsih me, H and red and Kay and blue. So for X, cubed over four is going to be kind of like an X cubed graph, but a little bit smushed. And kay of X is going to be like the X to the 1/3 graf. But also a little smushed just like this, so we can see the two intersection points we care about. So this is the point to come a to and then down here is the point negative too common Negative too. We&#x27;re also kind of note here that we see this really nice symmetry about the line y equals X. So here&#x27;s the line Wyffels X we have This is really nice symmetry that if you reflect across the line in these two graphs for, like, mirror images of each other, I&#x27;m just right in K pax. Here&#x27;s what I pictures look like now in part C were asked to calculate sin tangent lines. So how do we find slopes of changing lines? We need to find derivatives. So let&#x27;s find a tch prime of X and K prime of X. So the drifter that h of X to find these tangent lines is the dread of X cubed over for which is to react squared over four. So this is just, you know, your standard chain rule our soon MI to standard power rule calculation to find this derivative, the chain rule is going to come in when we want to find the derivative of K. And next we have this expression for X to the power of 1/3 which is kind of like a composition. Our outer function is the 1/3 power and the inner function is four acts. So how do we apply the chain rule? We take the derivative of the outer function and we plug in the inter function and we multiply by the derivative of the inter function. So for over three times four acts to the negative 2/3 Perfect. Now we need to actually find the slopes of tangent lines. So at the point to comma, too, these folks of the tangent line they&#x27;re going to be a tch prime of two will be the slope of the tangent line h of accent, too. So this is three times two squared over four, which is three times for over four, which is equal to three k prime of two. Again, just plug in tow. K prime 1/4 atyou meet 4/3 times four times two to the negative, 2/3. So for half 4/3 times eight to the negative 2/3. Now to evaluate. This is kind of a funky things Easiest way too right. It is as eight tow the 1/3 to the negative too. Because 1/3 times negative too is negative to third. So this exponents manipulation is valid. And the reason why we do this is because eight to the 1/3 is really easy. To write down eight to the 1/3 is just too cute. Brood of eight is to to the negative too. Two to the negative too is 1/4 and so we&#x27;re left with 1/3 is our final answer These two to the negative forces 1/4. So it&#x27;s going to cancel out with the floor. So this is the slope of the tangent line of K evac set X equals to So here we just found these two slopes of tangent lines Now for the very bottom at negative too common Negative, too. So once we come down to this other intersection point being calculated the same thing h prime of X equals negative too is going to be three again and k prime of negative two will be 1/3 the calculations going to come out the same. So we see that the slopes of the tangent lines down here are actually the same is as what they were at the point to comment. We&#x27;re almost done. I&#x27;m going to squeeze in party right here just so we can keep looking at our picture, were asked what The tangent lines are at the origin. So let&#x27;s start with the red curve. So the tangent line at the horror at the origin. We know this grass very well. It&#x27;s actually horizontal. H prime of zero is three times zero squared over for which is just zero. So the tangent line that the origin is the horizontal line. Why equals zero? But now for K of X, we know that the tangent line at the origin is going to be the reflection of the line. Why equal zero of this tangent line along this curve? Why equals X? So when we get this reflection, we see that we now reflect to a vertical tangent line. So this was the tangent line to H of X and the tangent line to KFX is going to be the reflection of this. Which horizontal line reflects the vertical line? So is the vertical line. X equals zero tangent to cave axe

Ankit Gupta · Answer

here we are given a polynomial function, and we need to find up. Were you of the polar function and given plane given function If a check equal to one divided by two x Q minus X that one divided by four No, in park. Okay, edge off zero. We put X equal to zero in the given polynomial function. Trouble Tzeitel. One divided were true zero Q minus zero with one divided way for villa que with also zero and time to one wayto Vero Vero man a 00 we get when a pawn four that the under four part E no part B that is head off Do we will put X equal toe in the given polynomial function You get head off to one divided Puerto No, que my left to plus one Nobody the way four now Simplify when divided Where to took you with it. My enough. You last one divided by four one by two times it that is four. My next to last one divided by four four managed to to left one divided by four. Now we add the fractions. We get it divided by four. Yes, one de breaded way for here. We make that denominator common toe. Take LCD. Get nine upon four. That&#x27;s the answer for this part. No. Come to the park. Si. We will find hedge off minus boom. Similarly, we will put X equal to manage to in the given polynomial function. One divided by two managed to put a power three minus man. It&#x27;s time to bless. One way for there is equal to mind. If one divided by two, mind you that they&#x27;ve minus it. Manage time managed toe that is plus two. That&#x27;s one divided way for one divided by two times minus said that is minus four. Blessed to breath, one divided way for Mina four plus two that if minus two plus one divided by four No, adding the fractions. We will make that denominator common. Fife, have you have by enough It by four Breath One way for we get my nothing plus one minus seven. Divided by four under four Part Steve minus seven. Way for no part B, you find at Joffe minus one. Here we will put X equals minus one. In the given polynomial function, Etch off minus one. No wonder about it. Back to you my life. Thank you. My left, my lifetime. One left Wonder where did? By four we get find a friend Cube that is minus one Mind off one divided by two My lifetime minus 100 plus one plus one divided by four. Take that common denominator We get managed to divided by four. Let&#x27;s four divided by four left one divided by four He get a job minus one. Quito minus two. Blessed for do that&#x27;s 13 divided by four Death tha answer for our D.

Clarissa Noh · Answer

it&#x27;s Claire&#x27;s. So when you read here so if you&#x27;re gonna use a closure, we have D over DX. Each of nuts over x X is equal to this is equal to the derivative of H times X minus H of X times one all over X Square for X is equal to two in their h, the derivative of two times two minus age of two times one all over two square, just equal to negative. Three times two minus four times one. We&#x27;re just plugging in the values that we got, then we get negative £5.

James Kiss · Answer

the hon g of h of x, the composition. Think of this as she of each of X. This allows you to see that our outer overall structures that G function in our input into X in the G function is the entire each function. So we think of RG function is to x and then our xar input is the H of X function which is excellent 30 plus x squared. Well, sex plus one. We can distribute too to simplify this to get two extra the third plus two x where looks to eggs us too R h of g of x composition. Think of this as hh of G of X. This allows us to see that were in putting the entire G function in for X into the H function. So now our outer structure, our overall structure is H function which is X to the third plus X squared X plus one In with all of those ex inputs, we&#x27;re gonna put the entire G function, which is to X. So we&#x27;re gonna replace every one of these axes with two x so to exe cute is too cute. Execute or a execute plus two x a square to spare X squared. It&#x27;s four x squared. It&#x27;s two x plus one

Nez Nikoo · Answer

Okay. Use equation. They&#x27;re telling us to find each of exit H of X is minus slow. So I put a minus 12 year and you see right away I can cancel this three this file submit. This becomes four. So now the equation is X minus four. Swear pickle. Two months. Yes. Okay, so now get a square root of each side. Get X minus four is speak well to plus minus and minus one becomes I four becomes 22 becomes too. I right and X is equal to four plus minus two I with many of these very straightforward. But when you have to know is if I have minus four, I really have a minus one and it four. And if you get a square root of this, what happens? You have I It&#x27;s a sign and four becomes too. So this is saying is to I think a little extra explanation again, but check. Thank you.

Problem

In 2012 , Hurricane Sandy struck the East Coast. …

Problem 120 Medium Difficulty

$h(x)=(x+2)(x-4)^{3}$
$$
\begin{array}{l}{\text { (a) Identify the } x \text { -intercepts of the graph of } h .} \\ {\text { (b) What are the } x \text { -intercepts of the graph of }} \\ {y=h(x-2) ?}\end{array}
$$

Answer

a. $x=-2,4,4,4$
b. $=x(x-6)^{3}$

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a. $x=-2,4,4,4$b. $=x(x-6)^{3}$

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Heather Z.

Caleb E.

Maria G.

Absolute Value - Example 1

Absolute Value - Example 2

Michael SullivanCollege Algebra

Catherine R.

Heather Z.

Caleb E.

Maria G.

a. $x=-2,4,4,4$
b. $=x(x-6)^{3}$

Michael Sullivan

College Algebra