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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.)
$ ( -5, 0 ) $, $ ( 5, 0 ) $
$f(x)=a\left(x^{2}-25\right), a>0$$f(x)=a\left(x^{2}-25\right), a<0$
Algebra
Chapter 2
Polynomial and Rational Functions
Section 1
Quadratic Functions and Models
Quadratic Functions
Complex Numbers
Polynomials
Rational Functions
Karthik C.
April 2, 2022
Ij7
Campbell University
McMaster University
University of Michigan - Ann Arbor
Idaho State University
Lectures
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In mathematics, the absolu…
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In Exercises 65-70, find t…
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for this problem will find to quadratic functions one that opens up one that opens down Who's grafts have the following ex intercepts. So since we'LL need two functions, let's go ahead and call the first one F Now, since the'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'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's one of the requirements in the instructions. So that'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'LL also need a problem that opens down work. So let's call this one G, and we can go ahead and take F and let's just multiply that by minus one. And the reason I'm doing the minus one here is because that'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't change, the ex intercepts is we can see here if you plug in negative five or five in CG. In either case, you'LL get minus twenty five because of this negative outside that'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's less than zero. This insurance that are grab opens down work, so that's our other function here. So X squared minus twenty five and negative X Square plus twenty five. And that's our final answer
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