How do i sketch the graph
a(x-A)^2(x-b)^3(x-c)(x-d)^4 .The graph crosses y axis at E
then there is a diagram
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How do i sketch the graph
a(x-A)^2(x-b)^3(x-c)(x-d)^4 .The graph crosses y axis at E
then there is a diagram
Image - TinyPic - Free Image Hosting, Photo Sharing & Video Hosting
Hey skg94.
If you have a function in this form, then it makes graphing a lot easier.
I don't know where the turning points will be between the points A and B and also for C and D but basically the curve will be > 0 between B and C, past D, and before A, and < 0 between A and B and also between C and D, It will be zero at all the points A,B,C,D since if you plug in that value for x you will get 0 since 0 * anything = 0.
As for where the turning points are (i.e. where the graph turns around from decreasing to increasing or increasing to decreasing), you need to use calculus for that.
But in terms of the graph you know where its positive and negative and you also know that it has to go through the y-intercept point E as well so all this information gives a rough sketch of the graph.
Did they cover how to get the turning points in your class or do you just need to draw a rough sketch?
Hello, skg94
How do I sketch the graph?
. . $\displaystyle y \;=\;a(x-A)^2(x-B)^3(x-C)(x-D)^4$
The graph crosses the y-axis at E.
Code:| E o | | - - o - - - -o- | -o- - - - o - - A B | C D |
The graph has x-intercepts at $\displaystyle x = A$ and $\displaystyle x = D.$
Since their factors have even powers, the graph is tangent to the x-axis.
The graph has x-intercepts at $\displaystyle x = B$ and $\displaystyle x = C.$
Since their factors have odd powers, the graph passes through the x-axis.
Therefore, the graph looks like this:
Code:| o * | * | - - o - - - -o- | -o- - - - o - - * A * *B | C* * D * * * * | * * * * | * * | * |