graph help

**lizard4** · June 21st 2007, 07:03 PM

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

**Jhevon** · June 21st 2007, 07:18 PM

Originally Posted by lizard4

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

$f(x) = y = 3x + 2$
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 $y = 0$

$\Rightarrow 0 = 3x + 2$

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

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

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

But why is this? Let's see.

For y-intercept, set $x = 0$

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

$\Rightarrow y = 2$

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

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

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