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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.)

$ \left(-\frac{5}{2}, 0 \right) $, $ ( 2, 0 ) $

Concave Up Equation: $1<0$Concave Down Equation:$-1<0$

Algebra

Chapter 2

Polynomial and Rational Functions

Section 1

Quadratic Functions and Models

Quadratic Functions

Complex Numbers

Polynomials

Rational Functions

Campbell University

McMaster University

Baylor University

University of Michigan - Ann Arbor

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for this problem. Let's find two different quadratic functions. Let'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'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's a quadratic. Let's just go ahead and foil this thing up. So first outer inner last. So we do the first terms. That's X Square, the outer terms 5/2 x, the inner terms that'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'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's one of our answers. Now let's go ahead and do a similar argument for G, but this time we'd like to ensure that G opens downward. So let's take the same graph as F, but let's go ahead and multiply the whole thing by a negative one. So when we do that, we'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'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's our final answer.

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