Expert: Paul Klarreich - 6/14/2006

Question
I am unsure how to find the tangent of y=1+x+x^2, when a. x=a, b. x-coordinate of -1, c. x-coordinate of -1/2, d. x-coordinate of 1. After getting these answers, I am suppose to graph the curve and the tangents. I don't know how to do this either. Can you help me with this?

Thank you
C J

Answer
Hi, C J,

You wrote:
Subject:  Fonding Slope of Tangents
Question:  I am unsure how to find the tangent of y=1+x+x^2, when a. x=a, b. x-coordinate of -1, c. x-coordinate of -1/2, d. x-coordinate of 1. After getting these answers, I am suppose to graph the curve and the tangents. I don't know how to do this either. Can you help me with this?

Thank you
C J
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For these problems, I recommend that you send me the EXACT instructions for the problem.  'find the tangent of ...'  is vague and, in a sense, meaningless, because you could be asked to:

a. Find the EQUATION OF THE TANGENT LINE TO THE GRAPH OF ....  at  x = -1
a. Find the SLOPE OF THE TANGENT LINE TO THE GRAPH OF ....  at  x = -1

etc.  Which is it?  You definitely have to know.  If you are not sure what the instructions are, you are unlikely to understand the problem.

For this problem, I am going to assume that the problem says:

Find the equation of the tangent line to the graph of  y = 1 + x + x^2,

a. At  x = a
b. At  x = -1
c. At  x = -1/2
d. At  x = 1.

The general scheme for all of these goes this way:

I. You plan to use the Point-Slope FORM of the equation of a line:
   y - y0 = m(x - x0)
II. To do that, you will need x0.  That is given in the example.
III. You will need  y0.  Substitute x0 into the original function.
IV. You will need m.  Find the derivative and substitute x0 into it.
........................
For Part a:  x0 = a.
y0 = 1 + a + a^2

dy/dx = 1 + 2x   <<---  used for other parts, too.
At x0 = a,  m = 1 + 2a.

Use the P-S Form:
y - (1 + a + a^2) = (1 + 2a)(x - a)

Now do some simplifying.
........................
For part b:  x0 = -1
y0 = 1 - 1 + 1 = 1
m = 1 + 2(-1) = -1

Use the P-S Form:
y - 1 = -1(x - (-1))

y - 1 = -x - 1
y = -x
........................
For part c:  x0 = -1/2
(you can do this now)
For part d:  x0 = 1
(likewise)

At the end, you will plot the points you got from parts b,c,d.  You will also:

A. Use your knowledge that  y = 1 + x + x^2  is a quadratic function, therefore a parabola.
B. Use the fact that the slope in part c  was zero.  (You got that, not I.)  Therefore, that is the vertex.
C. Draw the tangent lines, going back to your elementary school (high school) work on the graph of a straight line.
D. and finally, the parabola.

Graph sketching is WORK!  Don't expect that you can just knock it off in a minute or two.  This one is a 25-minute job.