How do you find the equation of the line tangent to the graph of #f(x) = x^2 + 2x +1#, with the point (-3, 4)?

1 Answer
Jul 18, 2015

The "how" depends partly on what tools you have in your mathematical toolbox. ("How" often depends on this.)

Explanation:

There are several alternative ways of presenting the following ideas to students. They all lead to the same answer, but the details of how to get there are different.

I will assume that you have not yet learned about the derivative and rules for differentiation.

We want the equation of the line tangent to #f(x) = x^2 + 2x +1#, at the point #(-3, 4)#.
We have a point, but we need the slope:

We can define the slope of the line tangent to the graph of #f(x)# at the point #(a, f(a))# by:

#m = lim_(hrarr0)(f(a+h)-f(a))/h#

(There is another definition available. It give exactly the same answer in every case. but let's stick to just one.)

So, the slope of the tangent in this particular problem is:

#m = lim_(hrarr0)(f(-3+h)-f(-3))/h#

We can try substitution, but we get #0#'s in both the numerator and the denominator. That form gives us no information about the limit. (That is why it is called an indeterminate form.)

So we have some work to do. At each step, we can try substitution again, but I won't discuss it after every line.

#m = lim_(hrarr0)(f(-3+h)-f(-3))/h#

# = lim_(hrarr0)(((-3+h)^2+2(-3+h)+1)-((-3)^2+2(-3)+1))/h#

# = lim_(hrarr0)((9-6h+h^2-6+2h+1)-(4))/h#

# = lim_(hrarr0)(h^2-4h+4-4)/h#

# = lim_(hrarr0)(h^2-4h)/h = lim_(hrarr0)(h(h-4))/h #

# =lim_(hrarr0)(h-4)#

# =- 4#

In the last step, we evaluated the limit by substitution.

The slope of the tangent line is #m = -4#

Equation of the Tangent Line

So the line goes through the point #(-3,4)# and has slope #-4#.

The line has equations:

#y-4 = -4(x+3)#

#y - 4 = -4x -12#

#y = -4x-8#

(I would guess that "the equation" means the slope-intercept equation.)

Differentiation Rules

the rules (formulas) for finding the derivative make the process much faster:

#f(x) = x^2+2x+1# has derivative:

#f'(x) = 2x+2#

so the slope of the tangent at #x=-3# is

#m = f'(-3) = 2(-3)+2 = -4#