{Use of Tech} Equations of tangent lines
b. Use a graphi...

Question

{Use of Tech} Equations of tangent lines

b. Use a graphing utility to graph the curve and the tangent line on the same set of axes.

y = −3x²+2; a=1

Accepted Answer

Welcome back, everyone. In this problem, for the given equation Y E equals 4 X2 minus 5, plot the curve on its tangent line at the specified value where X E equals -2 on the same rectangular coordinate system. To solve this problem, let's first try to plot a curve, and then we'll worry about how to find the tangent line. Now, notice just by looking at our equation, Y equals 4 X2 minus 5, we can tell it's a parabola. And from the equation, we can derive certain features. OK? So, 4 Y equal to 4 X2 minus 5. 1st of all, Comparing it to the general form of a quadratic equation, we can tell that the coefficient of the squared term is 4 or A is 4. The coefficient of the linear term X equals 0 because there is no linear term, and the constant C equals -5. Now what does this tell us? Well, for starters, notice that A, which is 4, is greater than 0. So that tells us that the parabola opens upwards. OK. In other words, our parabola is going to look like this? Next, if we look at our B term here, OK, which is 0, that can help us to find the vertex. Because at the vertex, OK, at the vertex of our parabola. The X term is going to be equal to negative B divided by 2A, so that's negative 0 divided by 2 multiplied by 4, which equals 0. So at the vertex X equals 0, and then when X equals 0, we know that the Y term is going to be 4 multiplied by 02 + 5, which equals 02 minus 5, my apologies. For multiplied by 02 minus 5, which equals -5. OK. So 05 is the vertex, OK? But in a box here. So we know where our parabola will open. We know what the vertex is. Now we need to plot and to do that we need a few points. So let's select a few X values around the vertex, OK? So let me set up a table here of X and Y values. OK. And we could start from -2, -2 -1, 01 and 2. Those are the X values that we could use, OK? And now, all we need to do is to find the wire values to help us to plot our points. Now remember, we already know that when X is 0, Y is -5. And to find the other values, all we need to do is to substitute the X value into the equation for Y. I'll leave that to you to do on your own. But when you do that, you should find that when X is -2, Y is 11, when X is -1, Y is -1. When it's 1, Y is -1, and when it's 2, Y is 11. So now we have all the points that we need to plot. So if we go to our graph, when X equals -2, Y equals 11. When X equals -1, Y equals -1, it's probably somewhere here. OK, when X equals 0, Y is -5. OK. When X is 1, Y is -1, and when X is 2, Y is 11, OK. So now we have our points. Now we can plot our curve by connecting them. We should find that we get our curve to look something like that, OK? So this would be the curve or the graph of Y equals 4X2 minus 5. Now let's move on to the next part of our problem to find the tangent line at the value of X equals -2. What are we doing here? Well, basically, we want to find the equation for the line that is going to touch or curve once where X equals -2. Let's think about what that means. First of all, how do we find the equation of a line? Well, recall. OK, put that in right here. Recall that the point slope form of a line. Point, slope, form of a line. OK. Says or is in the form Y minus Y1 equals M multiplied by X minus X1, OK. Where M is the slope of our line and X1 Y1 are the coordinates of a point on our line. So if we can find a slope of our line and we can plug in a point that's on our line, then we should be able to rearrange the point slope form to find the equation of our tangent line and use that to plot. Now, how can we do that? Well, we also know that the slope of our line is going to be equal to the derivative of our curve at the point. So if we can differentiate our Y equation, OK, and substitute the X value of the point where the tangent is, then we should get the slope of the line. So let's go ahead and try to do that, OK? So 4 Y equals 4X2 minus 5. The derivative of Y with respect to X is going to be equal to, to differentiate 4X2 we bring down the power 2 multiplied by the term. to the power of 2 minus 1, that's X2 minus 1 or X to the 1. And the derivative of -5 is 0 because that's a constant. So that means the derivative of Y with respect to X equals 8 X. So that means then. That at X equals -2, then our slope M, OK, will be equal to the derivative substituting X to be negative2. So that's gonna be a multiplied by -2, which equals -16. So now we have the slope of our line. Can we find a point on our line? Well, for remember that our tangent, our point, sorry, here, our point is right here. But we also know that it's gonna be the same point that's on the curve. And when X equals negative 2 on the curve, Y equals 11. This is the 0.-211. So we already have a point on the slur, the, the, the curve, OK? So in other words, we can say that X1 Y1. Is to 11. So that means X1 is -2, Y is 11. And so if we substitute that into the point slope form, then we're gonna get Y minus, OK, Y minus 11. Equals -16 multiplied by X minus -2 or + 2. If we go ahead and solve here, then we should get we're gonna get Y minus 11 equals -16 X minus 32, and if we add 11 to both sides, we get Y to be equal to -16 X minus 21. This is the equation of our tangent line. Now that we have the equation, we just need to find two points from our equation that can help us to plot. Now again, as we said, we already know a point on the line, negative to 11. So let's try to find another point on the line, OK, when X equals 0, OK, that will give us the Y intercept. And X equals 0. Then Y equals -16 multiplied by 0 minus 21, which is 0 minus 21 or 21. Thus, 0 minus 21 is on our tangent line. Now that we have that on our tangent line, if we go back to our graph, OK, let me just get rid of this circuit here. Oops. I know on our tangent line, we go, sorry, on our graph, we go to 0-21, which is somewhere down here. Now, we have two points and we can plot our tangent line. And when we do that, our tangent line should look something like that. Let me try it a little bit more accurately. Let me try to get it better and when we do that, our tangent line should look something like that. Therefore, this is the graph that shows the plot that shows the curve and the tangent line on the same rectangular coordinate system. Thanks a lot for watching everyone. I hope this video helped.

{Use of Tech} Equations of tangent linesb. Use a graphing utility to graph the curve and the tangent line on the same set of axes.y = −3x²+2; a=1

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{Use of Tech} Equations of tangent lines
b. Use a graphing utility to graph the curve and the tangent line on the same set of axes.
y = −3x²+2; a=1