# Slope of Tangent???

• Feb 9th 2009, 12:02 PM
Mudd_101
Slope of Tangent???
Hey, I really have no idea how to do this so help is greatly appreciated :)

Use the equation of the slope of a secant line to find the slope of the secant line passing through (2,4) on y = x^2, where h = 1.

I believe the equation of the secant slope is {f(a+h) - f(a)}/h

but again, not that sure..

thanks
• Feb 9th 2009, 12:31 PM
Henderson
Quote:

Originally Posted by Mudd_101
Use the equation of the slope of a secant line to find the slope of the secant line passing through (2,4) on y = x^2, where h = 1.

I believe the equation of the secant slope is {f(a+h) - f(a)}/h

This is true, but it may work better to think about what it means, rather than memorizing an equation.

If h=1, you want a point where the x is 1 more than the point you're starting with. (If h=2, you want a point where the x is 2 more, etc.)

The point you're starting with has $x=2$, so you want a point where $x=2+1=3$.

Then go back to your equation and plug in x=3:
$y=(3)^2=9$