Functions and Graphs

Find th...

Question

Functions and Graphs

Find the natural domain and graph the functions in Exercises 15–20.

f(x) = 1 − 2x − x²

Accepted Answer

Hello there. Today we're going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Identify the domain and sketch the graph of the function. JFX is equal to 8 minus 2X minus x to the power of 2. Awesome. So it appears for this particular problem, we're asked to figure out what the domain is, and we're also asked to sketch a graph of the specific function JFX. So we're trying to find the domain of JFX and create a graph for it. So now that we know what we're ultimately trying to solve for, let us quickly know once again that our given function states that J of X is equal to 8 minus 2X minus X to the power of 2. And let us now try to identify the domain. We need to note that when we look at our function that it is a quadratic function, which means that it will be a polynomial, and that will mean, because it's a polynomial, it is defined for all real values of X. So that means that the domain is all real numbers, so the domain will be parenthesis negative infinity positive infinity parentheses. So now we need to identify the key features of the graph that we need in order to properly plot this function of JFX. So once again, we need to figure out the graph of a quadratic function is a parabola. So as we should recall, since we are dealing with a quadratic function, that it will produce a parabolic curve, so it's going to be a parabola-shaped curve. So now we need to find the vertex of our parabola. So, a quadratic function in the form of A X to the power of 2, so A multiplied by X 2, plus B X plus C will have a vertex at. X equals negative B divided by 2A. Fantastic. So now we need to rewrite our function to isolate and solve for X eventually, but let us take a moment here. When we rearrange our function for JFX, we will find that JFX is equal to negative X to the power 2, so we can say that JX and let's move it down just a little bit. So we can say that our function of JF X. is equal to x 2 minus 2 X + 8 when we rearrange it to write it in our quadratic function or the parabolic form so it looks more familiar. We can see that the coefficients are in this particular case is A is going to be equal to -1. B is gonna be equal to -2. And C is gonna be equal to simply 8. So that will mean when we solve for X, we will find that X is equal to -2 divided by 2 multiplied by -1. So when we perform that calculation, we will find that it's going to be simply equal to -1. So now our next step is we need to find J of -1. So J of -1, meaning we're plugging in a value of -1 in for X is going to be equal to 8 minus 2 multiplied by -1, minus -1 to the power of 2. So when we plug that into our calculator, we will find that it's simply equal to 9. And once again, that's when we plug in a value of -1 in for X into our original function for JFX. Fantastic. So, therefore, without a doubt, the vertex is going to be at parentheses 1.9. So that will be our vertex. So I'll say V equals vertex is equal to parentheses 1.9. So now we need to find the X intercepts or the roots. You may hear them call you might hear them be called roots. So we're now trying to find the X intercepts or the roots. So in order to solve for the roots, we need to set J X equal to 0 and then solve for X, so we need to take negative X2 minus 2X plus 8 is equal to 0, since once again we're setting J X equal to 0. And like I mentioned before, we need to rearrange this expression to isolate and solve for X all by itself. So when we do that, we will find that X + 4 multiplied by X minus 2 is equal to 0. So that will mean that X is equal to -4 and X is equal to 2. Fantastic. So that means that our X intercepts are going to be parentheses -4.0 and parentheses 2.0. Fantastic. We are on a roll. So now we need to find our y intercept. So in order to find our Y intercept, we need to set X equal to 0. So J of 0, plugging in a value of 0 in for X, we will find that J 0 is equal to 8 minus 2 multiplied by 0 minus. 0 to the power of 2, fantastic. So 0 to 2. So when we evaluate that expression, you will find that it's simply equal to 8. So that will mean that our Y intercept is 0.8 in parentheses, fantastic. So now we can officially sketch our graph. So, going back to look at our blank graph that we have, that was provided to us by the prom itself, and we create a graph of our function based on the information we found together. We will find that we have our vertex, which is going to be at -1.9, which is represented by a yellow dot, because our yellow dot are all the coordinate points that we found together, and we will find that we have our X intercept at -4.0 and 2.0, and we have our Y intercept at 0.8. And we note that the parabola or curve, which is represented in blue, or parabolic curve, once again is in blue, is opened downwards because A is equal to -1. And the graph is downwards facing parabola that passes through the above points and will be symmetric around X equals -1. And that's it. We've officially solved for this problem. Hooray, we did it. So to quickly recap, We have created our graph, and then once again we need to note that our domain is parenthesis negative infinity, infinity. And that's it. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Functions and Graphs

Find the natural domain and graph the functions in Exercises 15–20.

f(x) = 1 − 2x − x²

Key Concepts

Natural Domain

Graphing Functions

Quadratic Functions

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