Question 1161462
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To find the *[tex \Large x]-intercept, which is the point *[tex \Large (a,0)] with *[tex \Large a] being the value *[tex \Large x] assumes when *[tex \Large y\,=\,0], substitute *[tex \Large 0] for *[tex \Large y] and solve for *[tex \Large x]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -4x\ +\ 3(0)\ =\ -12]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{-12}{-4}\ =\ 3]


Hence, the *[tex \Large x]-intercept is *[tex \Large (3,0)]


To find the *[tex \Large y]-intercept, which is the point *[tex \Large (0,b)] with *[tex \Large b] being the value *[tex \Large y] assumes when *[tex \Large x\,=\,0], substitute *[tex \Large 0] for *[tex \Large x] and solve for *[tex \Large y]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -4(0)\ +\ 3y\ =\ -12]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \frac{-12}{3}\ =\ -4]


Hence, the *[tex \Large y]-intercept is *[tex \Large (0,-4)]


Plot the two points, *[tex \Large (0,-4)] and *[tex \Large (3,0)], then draw a straight line through the two points.


 *[illustration graphByIntercepts.jpg]
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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