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# (a) Find the slope of the tangent to the curve $y = 3 + 4x^2 - 2x^3$ at the point where $x = a$.(b) Find equations of the tangent lines at the points $(1, 5)$ and $(2, 3)$.(c) Graph the curve and both tangents on a common screen.

## a) $8 a-6 a^{2}$b) $y=2 x+3$ and $y=-8 x+19$

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we are the function Y equals three plus four X squared minus two X to the third. Uh The graph of this function will be a curve. And the first thing, the first thing that we want to do is find a slope of the line that is tangent to this curve at the point X equals A. So the slope of a line, tangent to the graph of this function to the curve at the point X equals a will simply be derivative of this function evaluated at. So the slope of the tangent line Uh at a point a is equal to the derivative of the function at the .8. Well, the derivative of three plus four X squared minus two X to the cube derivative of three constant zero derivative of four X squared is eight times X. Uh minus the derivative of two X cubed is six X squared. And we have to evaluate uh this derivative, that's why prime, we have to evaluate it when X equals X. So for any number a two slope of the tangent line at x equals a will be eight times X a minus six times x squared or minus six times a squares. That is the slope of the tangent line to this function at the point X equals a. Next we're going to look at two different tangent lines to this function. One tangent line Uh is going to touch the graph at the .105. Uh So X one is going to be one, why one is going to be five. And if we want to uh equation of that tangent line, we're going to use going to use the point slope form of the equation of a line from algebra one why minus Y. One equals M At Times X -X one. So we are uh finding the equation of a line that is tangent to this craft to this curve at this point. Um So specifically when uh the X value is one and that's important because the X one and y one, we already know what they are. Uh But um the slope is going to be given by the UAE prime of a expression that we found to slope of the tangent line is going to equal the derivative of the function at that particular value of X. So the slope of the tangent line is going to equal eight times a or in this case uh A the value that you're plugging in for X. Okay, think of this as your x coordinate that gets plugged into the derivative uh your derivative um is eight times a minus six times a square, depending on what value X. Is, depending on you know, what is the x coordinate where the tangent line is touching the graph? Well, the x coordinate is in this case one. So we're going to plug in one where we see a so the slope of this tangent line is going to be eight times a eight times one -6 times a squared or six times 1 square. Uh and that's simple enough because one square it is one times six is six, so 8 -6 is too. And so we have the equation of this particular tangent line. Why minus y. 15 equals two, slope m two times x minus x. One of course X one is one. Doing just a little bit of algebra. We would distribute this uh multiplication here, we would get to x minus two and then we would add uh five to both sides. So not writing down those basic algebra steps, we get Y equals two X. Uh and like I said, this would have been uh -2 and then you add the five to both sides. So it's plus straight. And so the equation of a line that is tangent uh to our function wire backs at this point Is y equals two times x plus straight now, ultimately we want to be able to graph the function which you see here along with the two tangent lines. So since we already have one of the tangent lines, uh the equation of the line that would be tangent to the function Through this point is given by this equation, y equals two x plus tree. So let's go ahead and put that in Desmond's Y equals two, X plus trade. And we can uh zoom in a little bit on our graph, slide it over and zoom in a little bit. So we wanted to find the line that was tangent to the curve at the 0.1 comma five. And so if we bring this down just a little bit more uh Here is X equals one. Here is Y equals five. So here's the 50.1 comma five. So this blue tangent line Is tangent to our red curb. Our red function at the .1 comma five. Now we need to find another tangent line. The equation of another tangent line to the function. This time we're going to have detain geant line going through the point. Uh two comma three. So X one is 2, Why one is 3. So We are going to have a line passing through the point to common three. And uh this line will be tangent to the graph of our function. The equation of this new tangent line once again is given by the form y minus y. One equals m times x minus x one. We know that why one in this case um is three. And we're gonna have to calculate them real quick. And then that's going to be times X -X one or X -2. So let's go ahead real quick and calculate the m. Now, once again the slope detainment line is equal to the derivative of the function at that particular X coordinate. So derivative of this function. Why prime of a. Was eight A minus six A squared. So let's rewrite that since we can't really see it. Uh Why prime of a derivative of our our original function? Why at A is equal to uh eight a minus six A squared. So when X is too, since the tangent line is passing through this point, when to coordinate X coordinate is too, the slope of the tangent line is going to be the derivative of the Y function evaluated when excess too. So the slope of the tangent line is the derivative uh X equals two. So we've got to find Y prime of two. So we just simply plug into where you see this little A here eight times a eight times two -6 times a squared -6 times two square. Now, eight times two is 16. Subtract not two squared is four times six is 24. So 16 minus 24 or negative eight. So the slope of our tangible line is going to be negative. So now we just clean this up with a little bit of eligible y minus three equals distribute the multiplication by the negative eight negative eight times X is negative eight. X -8 times. Think of this as -2 is plus 16. Last but not least add to three to both sides. We get y equals negative eight X plus 19. So y equals negative eight X plus 19 is the equation of the tangent line that passes through this point. So the equation of a line that was tangent to our original function passing through this point is given by this equation. So let's uh put this equation uh into dez most Y equals negative eight X plus 19. Uh All right. So this green function is uh the line that is tangent to our red curve, our function At the point to common three. So if you look at the point excess too. Uh I'm sorry over here. If you look at the point X is too And y equals three. That's this point right here. So this green tangent line is detention line whose equation we just found and it is tangent to our original function, our red curve at this point. So here you have our original function graft and red. And you have to tangent lines the blue one and the green one.

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