00:01
In this problem, we are asked to find the slope of the tangent line to a given parabola at the point 1 3.
00:07
We can find the slope of the tangent line using either definition 1 or equation 2.
00:12
Here we see that we have definition 1, where we can say that the slope of our tangent line is equal to the limit as x approaches a of our original function minus our function evaluated at sum value a, divided by x minus some value a.
00:28
And here we define a to be the x value at the desired point where we are drawing our tangent line.
00:34
So in this case, we can say that a is equal to 1.
00:39
So here we have definition 1, where we plugged in the number 1 for a.
00:43
We plugged in our original function, and then we plugged in our function evaluated at the value of a.
00:49
So we can start to simplify, first simplifying our numerator.
00:55
And when we do that, we end up with the following polynomial divided by x minus 1.
01:06
And we can notice that we can simplify and factor this polynomial that we have on top.
01:14
So when we do that, we end up getting x minus 1 times 3 minus x up top, and that is divided by x minus 1.
01:24
So we can go ahead and cancel these two quantities, and then we end up getting that we have the limit as x approaches 1 of 3 minus x.
01:37
So when we go ahead and evaluate our limit here, we get that the slope is equal to 3 minus 1 or that the slope is equal to 2.
01:47
We can go ahead and repeat this using equation 2.
01:51
With equation 2, we have a similar approach for finding our slope where we're using a limit where h is approaching 0.
01:58
We see that we have our function, and we're plugging in a plus h for any value of x, and we're subtracting our original function evaluated at the value of a, and this is all divided by the variable h.
02:13
So in this case, similar to our definition 1, a is also going to be equal to 1, because it is also defined as the x value at the desired point where we're drawing our tangent line.
02:25
So just as we did with our definition one, we can go ahead and start simplifying.
02:31
Here we have our limit where we've plugged in all of the important values.
02:36
We have our function where we've plugged in a plus h.
02:40
And then we have our function evaluated at a.
02:43
And we found that this was equal to three above by simplifying what we have in the parentheses here.
02:49
So we can go ahead and begin to simplify by distributing and simplifying what we have in our numerator once again...