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'Tacton:FnM Each question Tort pointx Partial credit alloacd Shor aIl work . Make > table ofvaluss and graph: 4=rGiven the cquation ol thc parzbola: (-2)&#x...

Question

'Tacton:FnM Each question Tort pointx Partial credit alloacd Shor aIl work . Make > table ofvaluss and graph: 4=rGiven the cquation ol thc parzbola: (-2)' - (Zpls ) 4 Find the coordinates of Ile %ertcz pont(Zpts ) nilc (nc equalludi "the 9zi8 cl Innelr?(Zpts ) Find Ihe >-nterc-FC(Zpts ) Grapt the parboli:

' Tacton: FnM Each question Tort pointx Partial credit alloacd Shor aIl work . Make > table ofvaluss and graph: 4=r Given the cquation ol thc parzbola: (-2)' - (Zpls ) 4 Find the coordinates of Ile %ertcz pont (Zpts ) nilc (nc equalludi "the 9zi8 cl Innelr? (Zpts ) Find Ihe >-nterc-FC (Zpts ) Grapt the parboli:



Answers

Based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. Determine the point(s) of intersection of the graphs of $f(x)=2 x^{2}+7 x-4$ and $g(x)=6 x+11$ by solving $f(x)=g(x)$

And this problem We're trying to find the intersecting points between two functions FN. G. So I've written that out in the first step is to everything in one place we're going to add all of these terms over to the left. So that will give us three X squared plus 16 X plus five equals zero. So we're going to want to see if we can factor this. So the way I do it as I say three times 5 equals 15 And our middle time was 16. So we want to terms that multiply to get 15 but add to 16. So 15 times one I would add this turns you get 16. So we write that out. We had three X squared plus X plus 15 X plus five equals zero. And we're going to see what we can pull out. So these first two we can pull out X. We had three X plus one. Inside Plus We can pull out five. Get three X Plus one. So we get x plus five Times three x plus one equals zero. So now we just need to solve these Expose five equals 0 and three X plus one equals zero. So when we subtract five over we get X equals negative five. And then on the right we're going to subtract one and then divide by three. So we're going to get x equals negative 1/3. But we're not done with the problem yet. We still want to figure out what the Y values that correspond with these are. So it doesn't matter which functionally plug them into four and 85. Let's use -2 times x squared minus 11 X minus four. And with that we're going to get one So one point of intersection is negative five one. And for negative 1 3rd I'm going to go ahead and use the other function. It doesn't matter. You'll get the same answer. It's native one third squared plus five Times -1 3 plus one. And with that, you'll get native five nights. So you're two points of intersection are negative 51 negative one third negative five nights.

All right. So we're looking for the intersection points of two graphs. You can do this algebraic lee or graphene. If you're going to do it graphing, then the best way to do it is to make these equations equal to each other. Because what that means is they're going to have the same answer at the same X. Value, which is kind of the definition of an intersection point. So we make them equal. Then we want to try to solve this because it's a quadratic. We need to get it all to one side and find the zeros. So I'm gonna move my one equation over to my other side and make it equal to zero. So I'd have X square negative two, X plus three, X is a positive X And -4 -2 is a -6. From there. I can look at factoring. So I'm looking at what multiplies to negative six but adds to positive one. That would be a positive three and a negative too. Because my lead coefficient is one. I can just factor by inspection and then my answers here for X plus three, the one that makes that zero is negative three. And for X -2 the one that makes that zero is a positive too. So that tells me the two X values for my intersection points. And now all I have to do is substitute those back in to find my Y values. So it doesn't matter which equation you put them into because they should be the same answers. So negative three times negative three plus two. That works out to nine plus two which is 11. So one of my intersection points we should see at -3 and 11 And my other one make the three times two plus two, negative three times two is negative six plus two is a negative four. So we should see our other intersection point at two and negative four. So that would be the algebraic method for finding intersection points. We can also do it graphing uh for graphing. You'd want to put both of your equations in that you're solving for. So we are solving for a negative three X plus two as one of them. And we're solving for an X squared minus two X -4 for my other one. So there's our two. There would have to be able to see where the graphs cross. And then we would use our second trace intersect feature to figure them out. So there's the one we found. We said -3 and 11. We said -3 and 11 Algebraic Lee. And the other one down here we said we got negative or too negative for and al generically we said too negative for. So graphene or algebraic, you can get the intersex either way. Okay?

Okay, so we want to find that the intercept of this of dysfunction. Okay, so let let's first simplify the staff of facts. Okay? So for the top we have to we want to factor out And I into two. Ah two polynomial is multiple. E. Multiplied by each other. So it is easier for us to to cancel out, Right? So one is X plus three and then another one we have uh I think it doesn't work. Okay, It cannot be X plus three. Okay, It had to change. So here we have. Let's see -3. Okay. Plus two. Okay. Yeah it works uh and bottlenecks X plus three. Okay, so now we want to find the the intercepts of this graph of this function. So now let's find the X intercept first. Yeah. Here I have to define, don't remember. X cannot be -3. Okay. The domain effects cannot be ministry. Now let's find an excellent step so that the F. F. X equals to zero. Yeah, So we have two X -3 times X plus two over X plus three, it goes to zero. Okay, So we have x minus three. It goes to zero or X plus three, it goes to zero. Okay, So X could be 3/2, or X could be uh minus three. Right, Okay. So now let's find the Y intercept. Okay, so the Y intercept can be okay, So now I have to plugging actually goes to zero. Okay, So f of zero, it goes to nine and three times two Over three. Okay, so we have -2. So the Y intercept his minus two, and the X intercept are 33 halves and uh minus three.

All right, we are looking for the intercepts for the following rational equation. So keep in mind for your X intercepts Why is equal to 0? And in a rational equation, the only way you can get a. Y equals to zero is if the top is zero Because we can't divide by zero. So in this case are X intercept is at six. So my x intercept is 60 If I'm looking for a Y intercept then my X has to equal zero. So all I have to do is substitute in zero for X and solve it. So zero minus six is negative 60 plus two is two -6, divided by two is a -3. So I have a Y intercept At zero and -3. So our intercepts in this case are at 6, 0 & 0 -3.


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