WEBVTT
00:00:02.410 --> 00:00:10.600
Find the equation of the tangent to the curve π¦ equals eight π₯ squared plus five π₯ minus six at the point negative one, negative three.
00:00:12.380 --> 00:00:19.970
So, weβve been given the equation of a curve, it is a quadratic curve, and asked to find the equation of the tangent to the curve at a particular point.
00:00:21.430 --> 00:00:29.980
We recall, first of all, that the tangent to a curve at a given point passes through that point and has the same slope as the curve at that point.
00:00:31.210 --> 00:00:32.820
A tangent is a straight line.
00:00:32.820 --> 00:00:40.900
So, we can use the general form of the equation of a straight line, π¦ minus π¦ one equals ππ₯ minus π₯ one, in order to find its equation.
00:00:41.150 --> 00:00:47.930
Here, π represents the slope of the tangent, and the point π₯ one, π¦ one is a point on the line.
00:00:49.260 --> 00:00:57.290
We know that the point with coordinates negative one, negative three lies on this line because weβre looking for the tangent to the given curve at this point.
00:00:57.510 --> 00:01:06.060
And so, substituting negative one for π₯ one and negative three for π¦ one gives π¦ minus negative three equals ππ₯ minus negative one.
00:01:07.220 --> 00:01:11.410
This can of course be simplified to π¦ plus three equals ππ₯ plus one.
00:01:11.980 --> 00:01:15.400
Now, we need to determine the value of π, the slope of the tangent.
00:01:15.570 --> 00:01:19.520
And recall that this is the same as the slope of the curve at this point.
00:01:19.750 --> 00:01:25.330
We can therefore use differentiation in order to find the slope of both the tangent and the curve.
00:01:26.790 --> 00:01:35.160
π¦ is a polynomial function of π₯, so we can use the power rule for differentiation in order to find its derivative dπ¦ by dπ₯.
00:01:36.220 --> 00:01:42.750
We recall that to find the derivative of a power of π₯, we multiply by that power and then decrease the power by one.
00:01:43.100 --> 00:01:49.020
So, we have that dπ¦ by dπ₯ is equal to eight multiplied by two π₯ plus five multiplied by one.
00:01:49.230 --> 00:01:53.730
And then, the derivative of a constant, negative six in this case, is just zero.
00:01:55.000 --> 00:01:59.480
Our expression for dπ¦ by dπ₯ therefore simplifies to 16π₯ plus five.
00:02:00.670 --> 00:02:04.690
Now, this is a general function for the slope at any point on the curve.
00:02:04.900 --> 00:02:07.870
We need to evaluate it at the point weβre interested in.
00:02:08.060 --> 00:02:10.490
Thatβs the point where π₯ is equal to negative one.
00:02:11.730 --> 00:02:22.330
Substituting negative one for π₯ gives that the derivative of π¦ with respect to π₯ at this point is equal to 16 multiplied by negative one plus five, which is equal to negative 11.
00:02:23.650 --> 00:02:28.700
So, we now know the slope of the curve, and hence the slope of the tangent to the curve at this point.
00:02:28.990 --> 00:02:34.020
The final step is to substitute this value of π into the equation of our tangent.
00:02:35.320 --> 00:02:39.900
Doing so gives π¦ plus three equals negative 11 multiplied by π₯ plus one.
00:02:41.130 --> 00:02:47.050
We can distribute the parentheses on the right-hand side to give π¦ plus three equals negative 11π₯ minus 11.
00:02:48.220 --> 00:02:55.250
And then collect all the terms on the left-hand side of the equation to give π¦ plus 11π₯ plus 14 is equal to zero.
00:02:56.740 --> 00:03:02.970
So, by using differentiation to find the slope of the curve and hence the slope of the tangent to the curve at this point.
00:03:03.170 --> 00:03:06.660
And also recalling the general form of the equation of a straight line.
00:03:06.880 --> 00:03:17.890
Weβve found that the equation of the tangent to the curve π¦ equals eight π₯ squared plus five π₯ minus six at the point negative one, negative three is π¦ plus 11π₯ plus 14 is equal to zero.