y=ax^7+bx^6+cx^5+dx^4+ex^3+fx^2+gh+i
How many turning points (complete turns) does this graph have? How many x-intercepts? Y-intercepts? Does the left end go up or down? How about the right end?
I haven't been able to get this at all, I have no idea how to do that on my graphing calculator. Any help would be appreciated.
2006-10-16 16:26:55 · 2 answers · asked by Anonymous in Education & Reference ➔ Homework Help
Answers cut off the end my equation. It looks like this, like a pattern:
...dx^4+ex^3+fx^2+gh+i
2006-10-16 16:27:50 · update #1
Done on a TI-84 plus, Silver Edition.
It is not a parabola. X=0; Y=0
No X,Y intercepts
Two lines, about horizontal, starting at the X Axis; one in the right, positive quadrant, one in the left, negative quadrant.
The right one starts at about 1/2 positive X, and heads horizontal to the Y Axis, up the graph
The left line starts at about -1X and heads in a horizontal fashion down the graph.
I just plugged in all the terms and this is what was returned on my calculator.
2006-10-16 17:02:57 · answer #1 · answered by Anonymous · 0⤊ 0⤋
I don't remember enough to answer that question off the top of my head. I can direct you to a program which should be able to graph it for you. It's free and works well. The link is below... give it a try.
2006-10-16 23:36:00 · answer #2 · answered by Deirdre H 7 · 0⤊ 0⤋