Let f[x] equal 25 minus x squared.
Use the limit definition of the derivative
to find the slope of the tangent line
at negative five comma zero,
then determine the equation of the tangent line
at negative five comma zero in the form y equals mx plus b,
which is slope-intercept form.
So, one method to find the slope of the tangent line
at x equals negative five would be
to find the derivative function using the limit definition
of the derivative shown below,
and then evaluating the derivative function
at x equals negative five.
However, we can find the slope of the tangent line
in one step by substituting negative five for x
in the limit definition of the derivative,
which will give us f prime of negative five,
which will be the slope of the tangent line
at x equals negative five.
So, taking this approach,
we can find the slope of the tangent line in just one step,
and that's what we will do in this lesson.
The slope of the tangent line m
is equal to f prime of negative five,
which is equal to the limit as h approaches zero of...
In the numerator, we have f of the quantity
negative five plus h minus f of negative five.
In the denominator, we have h.
And now, to evaluate the functions in the numerator.
To evaluate f of the quantity negative five plus h
we replace x in f[x] with negative five plus h,
which gives us 25 minus the square of negative five plus h.
And then, we have minus f of negative five,
which needs to be in parentheses.
So, we have 25 minus the square of negative five,
and all this is still divided by h.
Now, we need to begin simplifying the numerator.
Let's begin by squaring negative five plus h.
To square negative five plus h,
we have two factors of negative five plus h.
When multiplying two binomials, we have four products.
One, two, three, and four.
Negative five times negative five is positive 25.
Negative five times h is negative five h,
giving us minus five h.
H times negative five is negative five h,
giving us minus five h.
And then, we have plus h times h,
which gives us plus h squared.
Notice how we do have two like terms.
Negative five h minus five h is negative 10h.
So, going back to the limit,
in the numerator we now have 25 minus, in parentheses,
the square of negative five plus h,
which we now know is 25 minus 10h plus h squared.
Then we still have minus the square of negative five
is positive 25, giving us 25 minus 25.
And again, this is still all over h.
There's another closed parenthesis here.
Now, we will clear the parenthesis here.
Because of the subtraction,
we need to either subtract each term,
or think of distributing a negative one.
So, distributing negative one,
we have negative one times positive 25,
which is negative 25, giving us minus 25.
And then, negative one times negative 10h is positive 10h,
giving us plus 10h.
And negative one times h squared is negative h squared,
giving us minus h squared.
And then, we have minus, but notice 25 minus 25
is equal to zero, all over h.
Also, notice how we have 25 minus 25 here,
which simplifies to zero.
So, now we have the limit as h approaches zero
of the quantity 10h minus h squared, all divided by h.
But notice how 10h and h squared have a common factor of h.
Let's factor out the common factor of h,
which gives us h times the quantity 10 minus h, all over h.
So, the reason for factoring the numerator
is now we have a common factor of h
between the numerator and denominator
that will simplify to one.
If this factoring was confusing,
another option would be to break this up
into two separate fractions
by dividing the monomials in the numerator by h.
Notice how if we have 10h divided by h
minus h squared divided by h,
we still have 10 minus h.
So, we have the limit as h approaches zero of 10 minus h,
but h is going to zero,
and therefore, this limit is equal to 10,
which indicates a slope of the tangent line
at the point at negative five comma zero is 10.
So, we know the slope of the tangent line is 10.
So, for the equation of the tangent line,
we know y is equal to 10x plus b,
we still need to determine b, the vertical or y-intercept.
Notice the point negative five comma zero
is not the vertical or y-intercept,
but we can use that point to determine b,
since negative five comma zero is a point on the line.
When y is equal to zero, x must equal negative five.
So, performing substitution,
we have zero equals 10 times negative five, plus b.
So, again, we've substituted zero for y,
and negative five for x.
This gives us zero equals negative 50, plus b.
Adding 50 to both sides to solve for b,
we now know b, the vertical or y-intercept is positive 50,
and therefore, the equation of the tangent line
at negative five comma zero is y equals 10x plus 50.
And before we go, let's verify this graphically.
The graph of the original function
f[x] equals 25 minus x squared is graphed in red.
The tangent line y equals 10x plus 50
is graphed here in blue.
Notice how the blue line is tangent to the red curve
at negative five comma zero, verifying our work is correct.
I hope you found this helpful.