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Graph of y=x^2

by Menos
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Menos
#1
Feb14-06, 04:16 PM
P: 5
Interesting feature of this graph. Consider 2 points on the parabola, I'll take (-2,4) and (4,16). By multipling the positive x values (2*4=8), you can get the y-intercept of the line from (-2,4) to (4,16). Proof: The line including (-2,4) and (4, 16) is written as y=2x+8. Thus, the y-intercept is 8. My question is why does this work? I've been trying to figure it out for a while, and I am completly stumped on this one... Any help would be greatly appreciated!
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0rthodontist
#2
Feb14-06, 05:45 PM
Sci Advisor
P: 1,253
First find the general equation for the slope b in terms of x1, y1, x2, and y2. Then use the fact that y1 = x1^2 and y2 = x2^2.

Actually it's not the product of the absolute value of the x values, it's the opposite of the product of the x values.
Integral
#3
Feb14-06, 06:19 PM
Mentor
Integral's Avatar
P: 7,315
Start from the 2 point formula for a line.

[tex] \frac {y - y_1} {x - x_1} = \frac {y_2 - y_1} {x_2 - x_1} [/tex]

The formula for your parabola is

[tex] y = x^2 [/tex]

So we can write

[tex] y_1 = x_1^2 [/tex]
and
[tex] y_2 = x_2^2 [/tex]

Use this information in the 2 point formula to get

[tex] \frac {y - y_1} {x - x_1} = \frac {x_2^2 - x_1^2} {x_2 - x_1} [/tex]

Note that the numerator on the Right Hand Side is the differenc of squares and can be factored to get

[tex] \frac {y - y_1} {x - x_1} = \frac {(x_2 - x_1) (x_2 + x_1)} {x_2 - x_1} [/tex]

Cancel like factors in the RHS
[tex] \frac {y - y_1} {x - x_1} = (x_2 + x_1) [/tex]

Now rearrange this to get

[tex] y - y_1 = (x - x_1) (x_2 + x_1)[/tex]
Simplify to get:
[tex] y = x (x_2 + x_1) - x_1 x_2[/tex]

Clearly you are correct for the simple parabola, in addition it can be seen that the slope of the line is the sum of the x coordinates.


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