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# Tangent slope as instantaneous rate of change

## Video transcript

the data for three points on a smooth function f is given in the table so let's actually graph these just so that we can visualize it a little bit so our vertical axis let's just call that y is equal to f of x and then we have our horizontal axis we have our horizontal axis this is our x-axis and we're going to skip some space here just because we immediately go to very high values of of f of x and i'll even skip some space here since we already go up to 6.5 all the way up to 7.5 so let me just show that i'm skipping some space here and we're going to start let's say this is 108 108 109 110 and let's say that this right over here is 6.5 that is 7 and then that is 7.5 and we can plot these points 6.5 comma 108.25 that will put us right over there 7 comma 109.45 that puts us right around there and then 7.5 110.15 puts us right around right around here and these are just three points on a smooth function f so we can visualize what that smooth function f might look like it might look something might look something like this try my best to so maybe it looks something something like this so our smooth function f might look something who knows what it looks like after this so that's this right over here is this is the graph of y is equal to f of x y is equal to f of x now let's try to answer their questions what is the average rate of change of f with respect to x as x goes from 6.5 to 7 7 to 7.5 and 6.5 to 7.5 so let's do them one at a time 6.5 to 7. our change in x our change in x here our change in x is 7 minus 6.5 which is equal to 0.5 and our change in y here our change in y here is equal to let's see we end up at 109.45 minus 108.25 which is 1.2 so our average rate of change over this interval is our change in y over our change in x or 1.2 divided by 0.5 let me write this down our change in y over our change in x is equal to 1.2 over 0.5 which is equal to 2.4 so this is equal to 2.4 now let's do the next interval our change in x once again is 7.5 minus 6.5 change in x is still 0.5 that's hard to read let me write that a little bit neater so our change in x our change in x is still 0.5 7 to 7.5 and our change in y our change in y let's see we end up at 110.15 we started at 109.45 so our change is 0.7 so it's 0.7 so our change in y over change in x our change in y over change in x is equal to 0.7 over 0.5 which is equal to 1.4 1.4 and so you can see that the slope of the secant line this is essentially the slope of the secant line between these two points so this one this one right over here let me try my best to draw it this one right over here you can see it's it's steeper than the second one then the second one then this secant line right over here then this secant line right over there and now let's find the slope the average rate of change i should say or the slope of the secant the average rate of change over this interval which is the same thing as the slope of the secant line between that point and that point so let's think about our delta x our delta x and maybe i'll do it up here so our delta x we're going from 6.5 to 7.5 so our delta x is equal to one and what is our delta y so our delta y our change in y is equal to let's see we end up at 110.15 we start at 108.25 110.15 minus 108.25 we increase by 1.9 1.9 so our change in y over change in x is equal to 1.9 over 1 which is equal to 1.9 1.9 fair enough and that's the slope of this secant line between this point and this point right over here slope of the secant line it looks something like that and you see that it's it's slightly less deep than the magenta secant and it's slightly more steep than the orin secant it's kind of in between the two now they ask us this final question use the average rate of change of f on this the larger interval from here to here which we already figured out that's 1.9 as an approximation for the slope of the line tangent to f at x equals seven so we're trying to approximate the slope of the line tangent to f at f equals seven so the line tangent might look something like this and we see at least visually it looks like that that little blue line we drew this one right over here does look like it's at least the way i've drawn it it looks like it's a pretty good approximation for the slope up there so we'll use this as the slope as an approximation for the slope of the tangent line at of at two f at x equals seven they say write an equation of the line to f at seven uh write an equation for the line tangent f at seven hundred nine point four five using point-slope form so it's going to be a line we're going to use this as an approximation for slope and it's going to be it's going to contain this line is tangent it touches to the curve at this point and so point-slope form just as a reminder it's just it's just another way of saying that every point on this line every point on this line x comma y if we were to find the change in x and change in y relative to this point right over here it's always going to have a constant slope so one way you could say is if you take an arbitrary point x y on this line you could say you could say your change in y so y minus 109.45 109.45 over your change in x so we know that the 0.7 is there or the 0.7 is is or the point x what we we can use x when when f of x when x is equal to 7 f of x is 109.45 so we're referencing the same point when our change in x is x minus 7 our change in y is y minus 109.45 and this is always going to be constant for any xy that i pick on this line there and we're using this as an approximation for what that's going to be that's going to be equal to 1.9 so if you want to put it in point slope form you just multiply both sides times x minus 7 and you get you get y minus 109.45 is equal to 1.9 times x minus 7. and we are done that is an approximation we've approximated the slope and if this was the slope this would be the equation of the tangent line in point-slope form