Graph each polynomial function. ƒ(x)=-2x^4+7x^3-4x^2-4x

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Hello, everybody. I read everything right today. Today, we're going to be looking at this question that states consider the given po polynomial function and plot its graph where our function is FX is equal to negative 17 X to the fourth power plus 304 X cubed minus 1341 X squared minus 162 X. Now, when looking at a question like this where the function, you know, has a lot of exponents, a lot of big numbers, you might be a little hesitant to go ahead and try to solve it. However, if we go to it little by little, you'll see, we, we will reach the answer at the end. So let's try to dissect this a little bit and to do that, I'll do some factoring. So I'll factor out first a negative X of our equation which will leave us with 17 X with a negative X multiplying an open parentheses with 17 X cubed minus 304 X squared plus 1341 X plus 162. And we close that parentheses. Now we can do a little bit of more factoring. So we'll have a negative X multiplying an open parentheses of 17 X plus two closed parentheses, multiplying open parentheses, X minus nine closed parentheses squared. So we've turned that big function that we had into something a little that looks a little less intimidating. We have two parentheses with an X month spang at the end and only one square. So now we can solve for the zeros of our graph. And to do that, we just set each X term equal to zero. So we have an X outside of every parent, the C. So we'll just have X equal to zero. Then we have a 17 X plus two equals zero. And then we'll have X minus nine, equal zero. So for 17 X plus two, we move the two over from the left hand side to the right hand side and we'll have a 17 X is equal to negative two. And if we divide both sides by 17, we'll have that X is equal to negative two divided by 17. Now, for the X minus nine, we'll just move the nine over from the left to the right and we'll have that X is equal to nine. So we'll have zeros at zero, comma zero, another one at negative two divided by 17 comma zero and another one at nine comma zero. So now we have three X intercepts next, we'll look to see for our Y intercept and to do that we'll just set our X equal zero. So we'll have F of zero equals negative M X which is in this case, zero, multiplying open parentheses of multiplying X which is zero plus two closed parentheses, multiplying a close parenthesis, zero minus nine or open parentheses, zero minus nine, close parentheses squared. Now, because we have a zero in front of all this, that's multiplying all of the terms whatever is inside of the parentheses won't matter because we're multiplying it by zero at the end and anything multiplied by zero is zero. So Y intercept will be at zero comma zero, which we could have determined from the fact that we have a zero at zero comma zero, we also have a Y intercept of zero comma zero. So now we will look at the different multiplicities of our graph. So X equals zero and X equals negative two divided by 17 have a multiplicity of one. And you determine this by looking at the exponent that each of those terms is associated with. So the X outside of the parentheses, which is where we got X zero, X X equals zero. That has an exponent of one. It has no other exponent and the negative two divided by 17 is associated with the parentheses of 17 X plus two. And that parentheses isn't raised to any other power except one. So they each have a multiplicity of one. Now X now what is, what does the multiplicity of one mean. So because we got a simplicity of one, well, therefore, we will cross X axis at each point. So now we look at our other zero, we had an X equal nine. So at X equals nine that has a multiplicity of two, which is an even number, which means that there will be a tangent at that point and we would change direction directions. So finally, the last thing I want to determine is how would each end of my graph look like? So how will the left hand side end and the right hand side of my graph end? Well, end of graph is determined by our dominant term or a dominant exponent which in this case is the 17 X squared or, or negative 17 X squared or X to the fourth, negative 17 X to the fourth power. That's the largest exponent we have in our equation. Therefore, ends will be both facing downward. So the left end will be down and the right end will be down. And now with all of this information, I can go ahead and draw a graph. So I draw a Y axis and, and X axis and we can plug in some points. So we know we have a point at zero, comma zero, we know we have another point at a very small. So let's say this is right negative one very far out because we have an X or a very small point of negative two divided by 17 comma zero, you have a point there and we had another point at nine comma zero. So we just draw nine and that point will be nine comma zero. Now our y intercept is also at zero comma zero. So now we just look to see the shapes of our graph. So we know each end will be down. So from the negative two divided by 17 comma zero to the left, our graph will be downwards and from the nine comma zero to the right, we will also be downwards. So now we have to see what happens between the negative two divided by 17 comma zero and the zero comma zero. Well, we saw that each of those points has a multiplicity of one, which means we cross the X axis at each point. So we cross the X axis at negative two divided by comma zero. And we go into above the X axis and then we have to change directions to be able to then intersect zero comma zero and intersect and cross the ax axis again and go below the X axis. And then we have to cross and or we have to change directions once again to be able to intersect the point at nine comma zero and nine comma zero. We had a multiplicity of two, which means it's a tangent. So we don't cross the X axis, we just touch the point and then change directions once again to below the x axis. So in the end, we should have a graph that kind of looks like an M. So I hope that was helpful. And until next time.

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Graph each polynomial function. ƒ(x)=-2x^4+7x^3-4x^2-4x

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