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View Full Version : How do I find the x-intercepts of a graph?


mathsucks12
May 29, 2008, 03:40 PM
:confused:

galactus
May 29, 2008, 05:07 PM
Set y=0 and solve for x.

AJ54
May 30, 2008, 12:52 AM
if it is a quadratic, factorise using adds to give b and multiplies to give ac. i.e. when y =ax^2 + bx + c. or just let y = 0.
hope this kind of helped!

kateuk
Jun 4, 2008, 12:00 PM
If it's a quadratic, then take the equation to be y = ax^2 + bx + c.
(Remember - if it simply says x^2, then the value of a is 1, NOT 0)
Then substitute into the following equation:
http://upload.wikimedia.org/math/3/e/a/3ea647783b5121989cd87ca3bb558916.png

This will give you two values of x, if you do it once taking the +/- as +, and the second time as a subtract sign.
This only really works if it's simple maths or you have a calculator. If not factorize it.
Then, the values of x are the numbers in the brackets multiplied by -1.
Here's an example:
x^2+6x+5 = (x+1)(x+5)
1 x -1 = -1
5 x -1 = -5

Therefore the x values are -1 and -5.


If it's a linear equation, for example, 2x+4=y
Take y to be 0.
So 2x+4=0
Then apply the following steps:
2x-4=0-4
2x=-4
2x/2=-4/2
x=-2
So, -2 is your intercept. With a quadratic equation there's 2 intercepts, with a linear only one. Does that all make sense? Eek, hope so!