Register to reply

Graph x^2-4x+3

by ocean09
Tags: graph, x24x
Share this thread:
ocean09
#1
Sep3-06, 02:32 PM
P: 16
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!
Phys.Org News Partner Mathematics news on Phys.org
'Moral victories' might spare you from losing again
Fair cake cutting gets its own algorithm
Effort to model Facebook yields key to famous math problem (and a prize)
Hurkyl
#2
Sep3-06, 02:40 PM
Emeritus
Sci Advisor
PF Gold
Hurkyl's Avatar
P: 16,092
Do you know what kind of shape the graph has?
ocean09
#3
Sep3-06, 02:52 PM
P: 16
Yes, a parabola.

shmoe
#4
Sep3-06, 03:09 PM
Sci Advisor
HW Helper
P: 1,995
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
#5
Sep3-06, 05:16 PM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,345
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
#6
Sep3-06, 06:31 PM
P: 16
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
#7
Sep3-06, 06:52 PM
Sci Advisor
HW Helper
P: 3,684
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
#8
Sep3-06, 08:12 PM
P: 16
b/c i want to know how to graph it
d_leet
#9
Sep3-06, 08:26 PM
P: 1,076
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
#10
Sep4-06, 02:02 PM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,345
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
#11
Sep4-06, 10:52 PM
P: 16
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.


Register to reply