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