Graph each polynomial function. Factor first if the polynomial is...

Question

Graph each polynomial function. Factor first if the polynomial is not in factored form. See Examples 3 and 4. ƒ(x)=2x^3(x^2-4)(x-1)

Accepted Answer

Hello. Today we're gonna be sketching the graph of the following polynomial function. So the polynomial function given to us is six X cubed times X squared minus one times X minus two. Now, in order to sketch the graph, we need to go ahead and first figure out the regions of increasing and decreasing on the graph. And in order to do that, we're going to need to first find the zeros of X. In order to get the zeros of X, we're going to set the polynomial function equal to zero. Then we want to go ahead and solve for each factor of X. So we need to fully factor the polynomial function. Currently, the polynomial function is in a factored form. However, the quantity X squared minus one can be factored further. We can factor X squared minus one by using the difference of squares. And by using the difference of squares, we can factor the polynomial to give us six X cubed times X plus one times X minus one times X minus two. And that is going to equal to zero. Now that we have the polynomial function in a fully factored form, we can go ahead and set each factor of X equal to zero by using the zero property that is going to give us six X cubed is equal to zero, X plus one is equal to zero, X minus one is equal to zero and X minus two is equal to zero. Now, we need to solve for each individual value of X. In the first equation, we can divide both sides of the equation by six. And that is going to give us X cubed is equal to zero. And if we take the cube root of both sides of this equation, we still get X is equal to zero. So X is equal to zero is going to be one of the zeros for the function. Next in the second equation, if we subtract both sides of the equation by negative one, we get X is equal to negative one. In the third equation, we just add one to both sides of the equation to give us X is equal to one. And finally, in the last equation, we add two to both sides of the equation. And that is going to give us X is equal to two. So we have four possible zeros for the number line in order to find the regions of increasing and decreasing. So let's go ahead and list them out on the number line. We're going to draw a number line that goes from negative infinity to positive infinity and the zeros that we calculated are going to be negative 10, positive one and positive two. So now what we want to go ahead and do is you want to take a testing point between each region of the graph. And if the value of that point produces a positive or negative number, we're going to use the sign of the number to determine whether that region is going to be increasing or decreasing. So there's one trick that we can do in order to figure out the increasing or decreasing of the end points of the graph in the original polynomial function, let's go ahead and add the leading powers of each leading X. So that is going to give us a power of three plus the power of two plus a power of one. And that is going to give us a power of six. Furthermore, we're going to change the exponent of the leading term to this new exponent and that is going to give us six X to the power of six. Now, whenever a number or the leading term has an exponent of an even number that tells us that the end points are going to be facing in the same direction. And the leading coefficient is a positive number. So that tells us that the end behaviors of this graph are gonna be that both sides of the end behaviors are going to be increasing to positive infinity. So the end points of the graph are going to be increasing. Now, we just need to take some testing points for the regions between negative one and 00 and one and one and two. Let's go ahead and take a testing point between negative one and zero. A possible testing point for that region is going to be X is equal to negative 0.5. Now, we can go ahead and plug in this testing point either to the original function or to the factored form of the function. If we plug this in to the original function, this is going to give us six times 0.5 cubed which is going to give us negative 0.75 times ze- negative 0.5 squared minus one, which is going to give us 1.25. And this is gonna be multiplied by 0.5, negative 0.5 minus two, which is going to give us negative 2.5 and multiplying these values together is going to give us an approximate value of negative 1.41. Now, since the value of this testing point is negative, that is going to tell us that the region between negative one and zero is going to be decreasing. Now let's go ahead and pick a testing point between zero and one. Well, that testing point can be X is equal to 0.5. If we plug it into the original equation, six times 0.5 cubed is going to give us 0.75 0.5 squared minus one is going to give us negative 0. 0.5 minus two is going to give us negative 1.5. And if we multiply these values together, we're going to get an approximate value of 0.84. Now, because this is a positive number, this tells us that the region between zero and one is going to be increasing. And finally, let's take a testing point between one and two. Well, that testing point can be when X is equal to 1.5. So six times 1.5 cubed is going to give us a value of 20.25 1.5 squared minus one is going to give us the value of 1. and 1.5 minus two is going to give us the value of negative 0.5. And multiplying these three numbers together is going to produce an approximate value of negative 12.66. And because this value is negative, this tells us that the region between one and two is going to be decreasing. Now, there's one more thing that we need to find and that's going to be the Y intercept. And for the Y intercept, we want to find a point that lies exactly on the Y axis. And this is going to be the region when X is equal to zero. So we're going to plug in zero to our original function. But notice that if we plug in zero to the original function, we're gonna be multiplying the outcome of these two terms by the value of zero. So this tells us that the value of F of zero is going to equal to zero, which means that the Y intercept is going to be located at the origin zero comma zero. So now that we have the regions of increasing and decreasing and the Y intercept, we can go ahead and make a rough sketch of what the graph is going to look like. So if we make a quick access, we're first going to label the zeros of X and that's going to be negative one at zero, which is also going to be the Y intercept at positive one. Then at positive two. And we know the end behaviors of this graph, we know that when the graph approaches from negative infinity to posit negative one, the graph is going to be increasing. And the same thing for 12 to infinity, the graph is going to be increasing. Now we know that the region between negative one and zero is going to be decreasing. So we can go ahead and draw a negative slope between negative one and zero, then the graph is going to increase between zero and one. So we're gonna have a positive slope between zero and one. And finally, the region between one and two is going to be decreasing. So we're going to have some type of negative hill between one and two. So the shape of the graph is going to look like a W and now we just need to pick the option above that accurately represents the information on this graph. Well, that option is going to be D and to show this, we have the zeros at negative 10, positive one and positive two. And the, the shape of the graph roughly matches the one that we came up with below. So that means that the answer to this problem is going to be D. So I hope this video helps you in understanding how to approach this problem. And I'll go ahead and see you all in the next video.

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Graph each polynomial function. Factor first if the polynomial is not in factored form. See Examples 3 and 4. ƒ(x)=2x^3(x^2-4)(x-1)

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