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Graphing Linear Functions in Standard Form

Graphing Linear Functions in Standard Form 1.2 - Solution

Each graph has a different set of $x\text{-}$ and $y\text{-}$intercepts. By finding these, we can identify which graph belongs to which equation.

$x\text{-}$intercept

The $x\text{-}$intercept occurs when $y=0.$ By substituting $0$ for $y$ into our equation and solving for $x,$ we can find the $x\text{-}$intercept.
$4x+3y=12$
$4x+3\cdot{\color{#0000FF}{0}}=12$
Solve for $x$
$4x+0=12$
$4x=12$
$x=3$
The $x\text{-}$intercept is at the point $(3,0).$

$y\text{-}$intercept

To find the $y\text{-}$intercept, we substitute $0$ for $x$ and solve for $y.$
$4x+3y=12$
$4\cdot{\color{#0000FF}{0}}+3y=12$
Solve for $y$
$0+3y=12$
$3y=12$
$y=4$
The $y\text{-}$intercept is at the point $(0,4).$

Graph

Now that we know the intercepts, we can draw the graph of this equation.

This graph corresponds to option $A.$