in-exercises-65-70-find-two-quadratic-functions-one-that-opens-upward-and-one-that-opens-downward--3

Question

In Exercises 65-70, find two quadratic functions, one that opens upward and one that opens downward, whose graphs have the given x-intercepts. (There are many correct answers.)

$ ( 0, 0 ) $, $ ( 10, 0 ) $

J Hardin · Accepted Answer

Let&#x27;s find a quadratic that has thie given x intercepts. So let&#x27;s try X minus zero and then x minus ten. And this method always works because you can check here if you plug in zero or if you plug in ten for X, then one of these two factors will be zero so that the whole function is there. So I guarantees that our function f does have the correct intercepts. And then if you just go ahead and simplify this and then distribute you see that leading coefficient is positive a positive one so that the graf opens upward. And similarly, we want a graph that also has the same X intercepts. But this time opens down work. So why don&#x27;t we just take our original function over here and just multiply this by a negative one? And you could you should commit yourself that if you plug in zero or ten and two g using the same reasoning is before we&#x27;LL get out zero because if you plug in zero for energy, then you get his unit over here. But excuse me, that should&#x27;ve been a minus. I gotta fix this in blue. And similarly, if you plug in ten for X, the second termer zero. So the function is zero. Now, when we go ahead and multiply out the eggs, this time we get a negative coefficient in the front. And so this tells you that the Graf opens down work, including the negative sign there and your answer, and these two make up our final answer.

J Hardin · Answer

for this one will find to quadratic functions. So remember what these look like. You write that down here. So here, this is where we want to write ffx a knot a one x So that&#x27;s a quadratic. And so we want one that opens upward and one that opens downward. And the grafts have the given intercepts. So if four comma zeros that intercept, then that it tells you that when he plugging for you get out zero. And similarly, when we plug in eight, we get zero. So that tells us that X minus four and X minus eight are both factors of the quadratic. So one examples we can just go ahead and write this and you can see that by writing f in this form. First of all, we can see that it&#x27;s a quadratic if you just go ahead and foiled us out. So this is the quadratic and we see the leading coefficient is equal to one which is bigger than zero. So this opens up word, so that will be one one of the functions. One of the quadratic ce. This&#x27;s the one that opens up word. And then now, to make this open on work. We could just go out and multiply this first function. So let me come back and use a different letter here. So let&#x27;s call this one g for a second. Quadratic. Let&#x27;s just go ahead and put a negative outside, and this will make it open downward without changing the intercepts. So here, once again, you can see that if you plug in for you get zero. And if you plug in a, you also get zero. So we still have the intercepts that we want given up here. And then if you go ahead and just foil this one out as well. So we see that the leading coefficient this time for this example is equal to negative one, which is less than zero. That means that it opens down work. And that&#x27;s what we wanted. We wanted two different quadratic ce with thes ex intercepts, one that opened up word one that open downward. And that&#x27;s your final answer.

J Hardin · Answer

for this problem. Let&#x27;s find two different quadratic functions. Let&#x27;s call them F and G. Those will be the the labels that we can use for our functions F and G one that opens up and one that opens down. And they have the following X intercepts. So this tells us that if you plug in negative 5/2 into FRG because it&#x27;s X intercept, that means the Y value has to be zero. And also, if you plug into into either refugee, why value is also zero the so the way we can do this is if we want X to have zero y value. When X is two or minus 5/2, we can do X minus two X minus negative 5/2 and you can see that if you plug in X equals two here or negative 5/2. In either case, one of the terms in the parentheses will be zero, so f will be zero at either one of these two points, and then also we see that this is a quadratic. So first, let me rewrite this as a plus sign and then go ahead and just to convince yourself that it&#x27;s a quadratic. Let&#x27;s just go ahead and foil this thing up. So first outer inner last. So we do the first terms. That&#x27;s X Square, the outer terms 5/2 x, the inner terms that&#x27;s minus two X. We can write Negative two is negative, 4/2, and then the very last term is minus two times 5/2. So that&#x27;s just minus five. And then we could go ahead and clean this up a little bit, combining the light terms so 5/2 minus 4/2 is 1/2. So this is a quadratic that has the route that X intercepts that are given. And since our positive are leading coefficient in this case is one. A equals one is bigger than zero, so this one opens upward. So that&#x27;s one of our answers. Now let&#x27;s go ahead and do a similar argument for G, but this time we&#x27;d like to ensure that G opens downward. So let&#x27;s take the same graph as F, but let&#x27;s go ahead and multiply the whole thing by a negative one. So when we do that, we&#x27;ll just have, which is the same as just putting the negative sign on this thing up here. So if you put a negative sign, But if you still plug into or negative 5/2 or just have negative zero, which is still zero, so the G will still have the same X intercepts. But now that we put the negative on the outside, this ensures that are leading coefficient will be negative. And that&#x27;s what we want for the graph to open downward. So Fergie we see that the leading coefficient is equal to negative one, which is less than zero. So this one opens downward and these are our two quadratic with the desired X intercepts and black F opens up and in green G opens down, so that&#x27;s our final answer.

J Hardin · Answer

for this problem will find to quadratic functions one that opens up one that opens down Who&#x27;s grafts have the following ex intercepts. So since we&#x27;LL need two functions, let&#x27;s go ahead and call the first one F Now, since the&#x27;s guys negative five and five zeros, this means f of negative five and F or five or both people say zero. So, in particular weaken do X minus five and then X minus negative five for the other route. And you can check here that if you plug in five, the first determined parentheses will be zero. If you plug in negative five, the second term and princesses will be zero. So this F that we wrote here will have these given X intercepts. And then let&#x27;s just go ahead and simplify this a bit and then go ahead and foil that out, and we see that the middle terms cancel. We definitely see that this is a quadratic degree to polynomial. That&#x27;s one of the requirements in the instructions. So that&#x27;s good. And we see here that the weeding coefficient is a equals one is bigger than zero, so that means that the Graf opens up work so this is one of our answers. However, we&#x27;LL also need a problem that opens down work. So let&#x27;s call this one G, and we can go ahead and take F and let&#x27;s just multiply that by minus one. And the reason I&#x27;m doing the minus one here is because that&#x27;LL take the leading coefficient and replace it with a negative number, and that I will ensure that the Graf opens down work. On the other hand, since all we did was multiplied by a negative number that won&#x27;t change, the ex intercepts is we can see here if you plug in negative five or five in CG. In either case, you&#x27;LL get minus twenty five because of this negative outside that&#x27;s not being square. And that plus twenty five, which is zero. So Chief Negative five and she a fiber, both zero. So that satisfies this, and we see that the leading coefficient now is minus one. That&#x27;s less than zero. This insurance that are grab opens down work, so that&#x27;s our other function here. So X squared minus twenty five and negative X Square plus twenty five. And that&#x27;s our final answer

J Hardin · Answer

Let&#x27;s go ahead and find a quadratic function that has the given X intercepts. So recall what these two are telling you. If we call fr quadratic, then if you plug in negative one or three and for X, the wine value should be zero. So why don&#x27;t we just go ahead and define? For example, there are many answers here that are correct. Let&#x27;s do X minus negative on and then times X minus three. And the reason for building this is that if you plug in either negative won in for X If you plug in that in the first factor zero if you plug in three, four ex than the second factor zero. So, in either case, FX will be zero. And let&#x27;s go ahead and just multiply this out. So first, let me rewrite this and then go ahead and foil this. Oh, and I don&#x27;t give you X squared minus two x in that ministry and you can see by the leading coefficient is positive one. So this craft opens upward. So that&#x27;s one answer here. Now what should we do if we want to Graft that opens down work? Well, one way we can do This is to just define a new function, Let&#x27;s say g of X and just multiply our original function by negative one. And just as before, you can check that G ofnegative one in G of three or both zero. So our graph will have these ex intercepts once again. But now, when we will supply this out. So basically, just multiply all of this by negative one. You see that the leading coefficient is negative. So this means that the Graf opens down work.

J Hardin · Answer

Let&#x27;s find a quadratic function F that has given X intercepts. So what we could do here is look at X minus Negative three x minus Negative a half and let me just go ahead and rewrite this. Now, The reason for doing this is that you can see that if you do. If you plug in negative three in sex than that, don&#x27;t make this factor zero so that the whole function is zero. And if you plug in negative one half that&#x27;LL make the second term zero so that the whole project is your own, and then to convince yourself that says this is a quadratic, you could go ahead and just multiply this so someone of foil this out and then just combined those X terms. And if you look at the leading coefficient, it&#x27;s a positive one. This means that the Graf opens upwards. So this is one of our answers, and now we would like another graph another problem. But this one that opens downward. So what we can do is just take our original function over here. It just won&#x27;t supplied by a negative one. This will ensure that the ex intercepts don&#x27;t change because once again, if you plug in negative three or negative behalf and for X, then one of these two factors will be zero. And then to just convince yourself that this is ah, Paraiba below that opens down where you could supply foil this out and multiplied by negative one. We&#x27;ve already foiled over here, so we could just multiply this all by negative one. And due to the negative sign on the front, we see that this one opens our

Problem

In Exercises 65-70, find two quadratic functions,…

Problem 67 Easy Difficulty

In Exercises 65-70, find two quadratic functions, one that opens upward and one that opens downward, whose graphs have the given x-intercepts. (There are many correct answers.)

$ ( 0, 0 ) $, $ ( 10, 0 ) $

Answer

$f(x)=a\left(x^{2}-10 x\right), a>0$
$f(x)=a\left(x^{2}-10 x\right), a<0$

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$f(x)=a\left(x^{2}-10 x\right), a>0$$f(x)=a\left(x^{2}-10 x\right), a<0$

Precalculus with Limits

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Alayna H.

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Absolute Value - Example 1

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Ron LarsonPrecalculus with Limits

Heather Z.

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Caleb E.

Michael J.

$f(x)=a\left(x^{2}-10 x\right), a>0$
$f(x)=a\left(x^{2}-10 x\right), a<0$

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Precalculus with Limits