Graph x^2-4x+3


by ocean09
Tags: graph, x24x
ocean09
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#1
Sep3-06, 02:32 PM
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Graph x^2-4x+3.

My question:

1) is -4x a slope?

I'm trying to graph this equation w/out using a calculator or plugging in points.


Thanks!
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Hurkyl
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#2
Sep3-06, 02:40 PM
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Do you know what kind of shape the graph has?
ocean09
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#3
Sep3-06, 02:52 PM
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Yes, a parabola.

shmoe
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#4
Sep3-06, 03:09 PM
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Graph x^2-4x+3


You might try completing the square, this should make it clear how it relates to the graph of the good old y=x^2.
HallsofIvy
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#5
Sep3-06, 05:16 PM
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No, "-4x" is not a slope. The slope of a line is a number. If y= mx+ b then the slope is the number m. In the example given, y=x2- 4x+ 3, the graph is not a line and so doesn't have a slope. One definition of "derivative" is that the derivative, at any given value of x, is the slope of the tangent line.
ocean09
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#6
Sep3-06, 06:31 PM
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Quote Quote by HallsofIvy
No, "-4x" is not a slope. The slope of a line is a number. If y= mx+ b then the slope is the number m. In the example given, y=x2- 4x+ 3, the graph is not a line and so doesn't have a slope. One definition of "derivative" is that the derivative, at any given value of x, is the slope of the tangent line.

How would you graph this equation w/out plugging in points or using a calculator?
CRGreathouse
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#7
Sep3-06, 06:52 PM
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Quote Quote by ocean09
How would you graph this equation w/out plugging in points or using a calculator?
Why would you want to graph it?
ocean09
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#8
Sep3-06, 08:12 PM
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b/c i want to know how to graph it
d_leet
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#9
Sep3-06, 08:26 PM
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Quote Quote by ocean09
b/c i want to know how to graph it
You know that it looks like a parabola. Are you familiar with the vertex form of the equation for a parabola? If so then doing what shmoe suggested would help you graph it.
HallsofIvy
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#10
Sep4-06, 02:02 PM
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Well, I think it was in fifth grade where they taught me to make a list of x y values:

x y= x2- 4x+ 3
0 3
1 1- 4+ 3= 0
-1 1+ 4+ 3= 8
2 4- 8+ 3= -1
-2 4+ 8+ 3= 15
3 9- 12+ 3= 0
-3 9+ 12+3= 24
etc, mark the points (0,3), (1, 0), (-1, 8), (2, -1), (-2, 15), (3, 0), (-3, 24) and then draw a smooth curve through the points.

Another method, that I think I didn't learn until 9th or 10th grade was to complete the square: if y= x2- 4x+ 3= x2- 4x+ 4- 4+ 3= (x- 2)2- 1. The graph is a parabola with vertex (2, -1) opening upward.
ocean09
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#11
Sep4-06, 10:52 PM
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Quote Quote by HallsofIvy
Another method, that I think I didn't learn until 9th or 10th grade was to complete the square: if y= x2- 4x+ 3= x2- 4x+ 4- 4+ 3= (x- 2)2- 1. The graph is a parabola with vertex (2, -1) opening upward.
i was looking through my math book, and it showed me how to graph it.

it doesn't hurt to sound stupid sometimes. at least, you are learning something.


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