<p>Can someone help me with sketching derivatives??? We will probably have to use msn so that I can show you the graphs.
The test is scheduled tomorrow!</p>
<p>Thanks.</p>
<p>Differentiate the equation and set it equal to 0. Those will be your x-intercepts
Take the second derivative and set it equal to 0. These will be your critical points (minimum/maximum)
i.e. Given y= x^3 - 3x^2 + 3
- dy/dx = 3x^2 - 6x
3x^2 - 6x = 0 (note: you can already tell it will look like a quadratic)
x = 0, 2
These are your roots/x-intercepts - Find the y-intercept of the derivative
3x^2 - 6x
3(0)^2 - 6(0)
y_int= 0 - Find the critical point (turning point/minimum) by taking the 2nd derivative and setting it equal to 0:
6x - 6 = 0
x= 1
So there’s a turning point at x=1
the y coordinate for that point is 3(1)^2 - 6(1) => 3 - 6 => -3
so the turning point of the parabola is at (1,-3) - Finally with the x intercepts, y intercept and the turning point you can sketch the parabola, which is the derivative of the original equation.</p>
<p>Hope that helps =)</p>
<p>Thanks. I figured everything out yesterday by myself.</p>
<p>oh oops
i’m sure you did great on your test =)</p>
<p>Thank you!</p>
<p>Graphing derivatives isn’t too bad if you know a little about the basic appearance of function graphs. What I sometimes have difficulty with is relating a derivative graph to the original or vice versa, without an equation given.</p>
<p>Nevermind. I read the first post wrong. Ignore this.</p>
<p>Also important are the relationships between increasing and decreasing and positive and negative. Those show up on FRQs a lot.</p>