graph help

Quote:

Originally Posted by lizard4

i need help with graphing (f)x = 3x+2
thank you

$\displaystyle f(x) = y = 3x + 2$
This is just a line with slope = 3 and y-intercept = 2

To draw a line, all we need are two points, the intercepts are always good points to use.

For x-intercept, set $\displaystyle y = 0$

$\displaystyle \Rightarrow 0 = 3x + 2$

$\displaystyle \Rightarrow x = - \frac {2}{3}$

So our x-intercept is $\displaystyle \left( - \frac {2}{3}, 0 \right)$

For y-intercept, we simply have the lone constant as the intercept. So our y-intercept is at $\displaystyle y = 2$, which is the point $\displaystyle (0,2)$

But why is this? Let's see.

For y-intercept, set $\displaystyle x = 0$

$\displaystyle \Rightarrow y = 3(0) + 2 = 2$

$\displaystyle \Rightarrow y = 2$

So the y-intercept is $\displaystyle (0,2)$

Now just plot these points on a graph and draw a straight line through them, as you see below