We want to find the point
where the tangent line to y equals x cubed
at x equals 0.8 crosses the x-axis.
So we're looking for the x intercept of the tangent line
to the function y equals x cubed when x equals 0.8.
So let's first look at this graphically.
In blue we have the graph of y equals x cubed.
When x equals 0.8, this would be the point on the function,
and therefore this red line is a tangent line at that point.
And our goal here is to find the x intercept
of the tangent line.
So we want the coordinates of this point here.
So to do this, we'll first find the point of tangency
which gives us a point on the line
and then to find the slope of the tangent line,
we'll evaluate the derivative function at x equals 0.8.
Once we find the equation of this tangent line,
to find the x intercept, we'll set y equal to zero
and solve for x.
So we'll begin by determining the point of tangency,
which, again, would have an x coordinate of 0.8,
and the y coordinate would be y of 0.8.
Notice how we have x as a decimal,
however since 0.8
is equal to 8/10, which simplifies to 4/5,
we could also express the point of tangency as
having an x coordinate of 4/5
and a y coordinate of y of 4/5.
So it shouldn't matter whether we use decimals or fractions,
however, if we do use decimals,
we definitely don't want to round,
because we don't want approximations for our coordinates.
So I think I'll go ahead and use fractions
to solve this problem.
So now we need to find y of 4/5.
Well y of 4/5
is equal to 4/5 to the third,
which is equal to four to the third
divided by five to the third,
which is equal to 64 divided by 125.
So the point of tangency, which is a point on the line,
would have an x coordinate of 4/5
and a y coordinate of 64 over 125.
So now to find the equation of our tangent line,
we need to find the slope which we can determine
by evaluating the derivative function at 4/5
or if we want, 0.8.
So, the slope of our tangent line
is equal to y prime of 4/5,
where y prime is equal to three x to the second,
and therefore y prime of 4/5 equals three
or three over one times 4/5 squared,
which would be three over one times 16/25,
which equals 48/25.
Now we have enough information to determine
the equation of our tangent line.
If we know it contains the point 4/5 comma
64 divided by 125,
and also has a slope of 48 divided by 25.
So using the point slope formula of a line,
the equation of the tangent line would be
y minus y sub one,
y sub one is 64 divided by 125
equals m or the slope, which is 48 divided by 25
times the quantity x minus x sub one,
x sub one is 4/5.
So this is the equation of our tangent line.
Now, of course, if we wanted to, we could solve this for y
and write it in slope-intercept form,
but that's not our goal here.
Our goal here is to find the x intercept.
So to find the x intercept of any function,
we set y equal to zero and solve for x.
Let's go ahead and do this on the next slide.
So if we set y equal to zero,
we would have the equation zero minus 64 divided by 125
or negative 64 divided by 125
equals 48 over 25
times the quantity x minus 4/5.
Now remember our goal here is to solve for x
to find the x intercept.
So while we could distribute, it'll be easier
if we multiply both sides of the equation
by the reciprocal of 48/25,
which would be 25 divided by 48.
So multiply both sides of the equation by 25 divided by 48.
Notice how on the right side,
these are reciprocals, so this product is just one.
So on the right side we just have x minus 4/5.
On the left side, notice how this simplifies.
There's 125 and 25, and five 25s in 125,
and 48 and 64 share a common factor of 16.
There are three 16s in 48,
and four 16s in 64.
But notice how we do have a negative here,
so on the left we have negative 4/15.
So the last step to solve for x
we would add 4/5 to both sides.
So we'd have negative 4/15
plus 4/5
equals x minus 4/5 plus 4/5.
Of course, on the right this simplifies to zero,
on the left we have to obtain a common denominator
which would be 15,
Multiply 4/5 by three over three,
so we have x equals, here we'd have negative 4/15
plus 12/15 which would be 8/15.
So this gives us the point where the tangent line
crosses the x-axis when x equals 0.8.
And there are a couple ways to express this.
We can say the x intercept is the point
where the x coordinate is 8/15
and the y coordinate is zero.
Or we can also say the x intercept
equals 8/15.
And I think in the homework you're asked to express
the x intercept in this form here.
So if we go back to our graph just for a moment,
we found this point of tangency to have coordinates
4/5 comma 64 divided by 125,
and we found the x intercept here
to have an x coordinate of 8/15,
and, of course, a y coordinate of zero.
So this graph does verify our work is correct.
I hope you found this helpful.