How do you graph the line 4x+3y-12=0?

Aug 4, 2016

$y$ intercept of the function is present at the point: $\left(0 , 4\right)$
and $x$ intercept of the function is present at the point: $\left(3 , 0\right)$
See full explanation for more detail:

Explanation:

In order to clearly recognise the $x$ and $y$ intercepts of the function we may convert the given function to linear gradient form. That being:
$y = m x + c$
Where:

$c$ is the constant determining the $y$ intercept.
and $m$ is the gradient of the function.

Therefore, let us do this:
$\to 3 y = - 4 x + 12$
Dividing both sides by $3$, we get:
$y = - \frac{4}{3} x + 4$

This implies that the $y$ intercept of the function is 4.
We can prove this via the statement:
$y$ intercepts where $x = 0$:
Therefore:
$y = 0 \cdot x + 4$
This implies that the $y$ intercept of the function is present at the point: $\left(0 , 4\right)$

We can thus determine the $x$ intercept using the statement:
$x$ intercepts where $y = 0$:
$\to 0 = - \frac{4}{3} x + 3$
$\therefore \frac{4}{3} x = 4$
$x = 3$
This implies that the $x$ intercept of the function is present at the point: $\left(3 , 0\right)$

If we plot these points on a Cartesian plane and draw a line between the two, the graph has been drawn.

Attached below is a graph of the function with the intercepts in frame:

graph{y=-4/3x+4 [-9.54, 10.46, -2.92, 7.08]}