Thread: Finding the slope/equation of the tangent line

No clue how to do these, missed class(sick ughhhh) and I'm trying to catch up on the example problems since we have a test in a couple weeks. If I know how to do these I can basically apply them to the rest in order to figure out how to do this. Anyways, here's the problems, big thanks in advance for the help.



Find the slope of the tangent line to the given curve at the given value of x. Find the equation of each tangent line.

1.) y = x^2 - 6x ; x = 2


2.) y = e^x ; x = 0


3.) y = ln x ; x = 1

Originally Posted by Mathamateur

Find the slope of the tangent line to the given curve at the given value of x. Find the equation of each tangent line.

1.) y = x^2 - 6x ; x = 2

y' = 2x - 6

At x = 2:
y = (2)^2 = 4

y' = 2(2) - 6 = -2

This is the slope of the tangent line, so we may write:
y = -2x + b
for the tangent line.

Now the tangent line passes through (2, 4):
4 = -2(2) + b

b = 8

Thus the tangent line at x = 2 is
y = -2x + 8

-Dan

Originally Posted by Mathamateur

Find the slope of the tangent line to the given curve at the given value of x. Find the equation of each tangent line.
2.) y = e^x ; x = 0

One more.

y' = e^x

At x = 0:
y = e^(0) = 1

y' = e^(0) = 1

Thus the tangent line is:
y = x + b

This line passes through (0, 1):
1 = 0 + b

b = 1

Thus the tangent line is:
y = x + 1

-Dan

Ok thanks

For the 3rd one I got the slope = 1

Then the line passing through the point (1,0)

so the equation of the tangent line would be y = x - 1

Am I right?

Originally Posted by Mathamateur

Ok thanks

For the 3rd one I got the slope = 1

Then the line passing through the point (1,0)

so the equation of the tangent line would be y = x - 1

Am I right?

Looks good to me!

-Dan