SOLUTION: 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