5

Select the correct graph of the function fand evaluate Iim fx), if it exists. r+ifx < 1{x) =ifr=]-x+ 2 ifx> 1...

Question

Select the correct graph of the function fand evaluate Iim fx), if it exists. r+ifx < 1{x) =ifr=]-x+ 2 ifx> 1

Select the correct graph of the function fand evaluate Iim fx), if it exists. r+ ifx < 1 {x) = ifr=] -x+ 2 ifx> 1



Answers

Match the function to the graph. $f(x)=-(1-x)^{2}+1$

In this problem we're going to take the function F. Of X equals negative X squared minus one. And see what it looks like when graphed. I'm going to start with the parent function that's just X squared. This is called a quadratic because we have too far exponents. When I graft this quadratic, I get a parabola. A parabola. Is this U. Shape here? Notice it's opening upward. It's got a vertex at 00 at the origin. And then I can change things in my function in my equation and it will also change my Parabola. The first change I noticed and this one is there's a negative on front of my ex. So whenever I do that in my equation and graph it it reflects my parabola over the X. Axis and now it's opening downward. My vertex is still at the origin. That hasn't changed but everything else has been reflected over. And then the second thing that's done in this equation is this minus one at the end. So what is going to happen once it's been reflected over the X. Axis and then we subtract one is it's gonna move my parabola down one unit on my coordinate plane, so I'm going to get rid of these other two. This is the graph that we're looking for for this function are Parabola opens downward and we've got a vertex that is also are Y intercept at zero negative one. And notice we do not have any X intercept intercepts for this graph.

Trying to match the graph of the function F of X equals X plus one. For the domain of X is less than to what you're looking for is an open circle at two, and then the graph to the left of it. This matches graf. See.

In this problem we're going to take a look at the function F. Of x equals negative the quantity of X plus one squared. And we're gonna start with our parent function which is just X squared. And we call this a quadratic because our exponents on the variable is a. Two. So whenever I graph my parent function it's a parabola. Notice that it is opening upward and it has a vertex at the origin 00. I can change pieces of my equation that will also make changes to my parabola. So the first change I'm going to make is I'm going to put the negative sign on the front of my variable. Once that's done, my parabola is reflected over the X. Axis, notice my vertex is still after the origin and then the last piece that I'm gonna do is we're going to add 12 X. Inside the parenthesis. It's very important that the plus one is inside the parenthesis. I keep the negative on the front. Whenever I do the addition of one. Inside the parentheses, it moves my entire parabola, one unit to the left, so I want to go out and get rid of these first two. This is the correct graph of our function and notice that the vertex is at negative 10 It is also um are X intercept and then we have a Y intercept at zero negative one.

In this problem we're going to take a look at the function F. Of X equals negative. The quantity one minus X squared. And we're gonna start with our parent function X squared. And see how it changes while building the larger function. So when I start with X squared. Which is called a quadratic because of the two for the exponent. I get this U. Shaped graph that we call a parabola and I can make changes to my function and it will change my parabola. So the first change I'm gonna do is the negative X. And when I put a negative in front of the X, it reflects my parabola over the X. Axis. And then I'm going to in the original problem there's actually two negatives. There's one in front of the parentheses and one inside the parentheses. So I wanted to put parentheses around that X. And add a negative in the front. And this actually doesn't change this part of my parabola because whenever you square a negative it just gives you the positive. So it doesn't change it. Um This negative here is not changing the shape. This one on the outside is doing the reflecting over the X. Axis. And then the last piece that we have here is we have a one minus the X. Inside the parentheses. So that is going to move my Parabola one unit to the right and I'm going to get rid of all of these extra ones. So this is the graph that we're looking for this parabola here that's opening downward and it has a Y intercept at zero negative one. And the vertex is also the X intercept at 10


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