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(a) Draw the graph of the parabola. (b) From your graph, estimate the $x$ -intercepts. (c) Check your estimate by finding the $x$ -intercepts exactly.$$y=-2(x-3)^{2}+5$$

$$\text { (b) } 1.4,4.6(\mathrm{c}) \frac{6 \pm \sqrt{10}}{2}$$

Algebra

Chapter 1

Functions and their Applications

Section 4

Quadratic Functions - Parabolas

Functions

Missouri State University

Campbell University

Oregon State University

McMaster University

Lectures

01:43

In mathematics, a function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. An example is the function that relates each real number x to its square x^2. The output of a function f corresponding to an input x is denoted by f(x).

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(a) Draw the graph of the …

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All right, so we need to craft dysfunction. We've got a lot of transformations happen. X minus three squared, uh, plus five. So what I have my students do is they have to show that we are shifting right? Three. It's always backwards. And the reason why it's backwards, as you can imagine, plugging it three and for X three minus three would give me zero squared. So that's where my Vertex will be located. Yeah, and then up. Five. That's from this number ships out. And that makes perfect sense to me. Uh, and then this negative is going to make this problem open downward. It's actually the X axis reflection. You know, some people say it's a downward. You depends on how your teacher craziest things. But there's also this vertical stretch of two by a factor of two. So what that means is, instead of this parabola opening upwards like it normally does, it shifted right, three up, five. That's where your Vertex is located. And then it's wounded. Open downward. And I like to think of this as like the first one opens up by two. And then the next one opens up by, uh well it's not going to be consistent by two. It will be more than that. Right now. I'm just drawing a blank about how much it opens up by, um, so all I can answer for this is what your graph looks like without a graphing calculator. All I can tell is that the answer for B is that your X intercept one of them is bigger than one, but smaller than two. I think they want you to write 1.4 there, and then the other 1123 bigger than four was smaller than five. Um, and by the way, the Vertex goes through the axis of cemetery as well. So we're to have that X equals three. So if I went to the left 1.6, I need to go to the right 1.6. So that's where you would get 4.6 for good estimates. And a graphing calculator might be better as well. Now, to get the best answer for your X intercepts, it's when y equals zero. So what you can do is set that equation equal to zero. Your first. It's pretty straightforward. You just subtract five over, cancel that out and also divide by negative two. So, as I'm looking at, my next step would be the X minus three squared would equal 5/2. But then when you go to square root, you don't want to leave an irrational number in the denominator so you can multiply top and bottom by the, uh, whoops route to or or two. So we're looking at plus or minus Route 10/2 times two is four, but you square for us to. So then your last step would be to add this three over. But what people usually do is they get the same denominators, so they would change that. To be six halves 65 by two equals three, so you can put that into the numerator. So the best answer for your X intercepts would be six plus a minus. Route 10 all over to I don't skip did two steps at once there at the end, but I hope that made perfect sense what I did

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