SOLUTION: Answer Checking: What is the slope of the tangent line through (0,-3) and (x,y) for y = 3x² - 3 and x=1? my answer is m=3x^2 but i think their looking for non variable answ

Algebra ->  Decimal-numbers -> SOLUTION: Answer Checking: What is the slope of the tangent line through (0,-3) and (x,y) for y = 3x² - 3 and x=1? my answer is m=3x^2 but i think their looking for non variable answ      Log On


   

Question 1121247: Answer Checking:
What is the slope of the tangent line through (0,-3) and (x,y) for y = 3x² - 3
and x=1?
my answer is m=3x^2
but i think their looking for non variable answer
Found 2 solutions by solver91311, Edwin McCravy:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


Why are you calling the line a tangent line? It isn't tangent to anything in your problem. You have two points, one given and one you have to calculate, and you want the slope. The given point is . If , then when , so your second point is .

Just use the slope formula:



Where and are the two points on the line. You can do your own arithmetic.


John

My calculator said it, I believe it, that settles it

Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
What is the slope of the tangent line through (0,-3) and (x,y)
for y = 3x² - 3 and x=1?
It's best to draw a graph when you can.



You want the slope of the green tangent line which goes through at (0,-3)
and the blue tangent line which is through the point where x=1, which has
the y-value found by substituting 1 for x in y = 3x² - 3
                        y = 3(1)² - 3
                        y = 3(1) - 3
                        y = 3 - 3
                        y = 0

So the blue line goes through the point (1,0). 

The derivative is a formula for finding the slope of a tangent line at any
point.   So we find the derivative:

                       y = 3x² - 3
                      y' = 6x

To find the slope of the green tangent line, we substitute the x value of
the point (0,-3), which is 0, in the derivative:

                      y' = 6(0)
                      y' = 0

So the slope of the green tangent line is 0. [You expect that because the green
line is horizontal, and any horizontal line has slope 0.


To find the slope of the blue tangent line, we substitute the x value of the
point (1,0), which is 1, in the derivative:

                      y' = 6(1)
                      y' = 6

So the slope of the blue tangent line is 6.  The blue line goes up to the right
so we would expect it to have a positive number for its slope, and a slope of 6
is pretty steep.

Edwin