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# Graphing Linear Functions in Standard Form

## Graphing Linear Functions in Standard Form 1.8 - Solution

a
We will graph this equation by finding and plotting its intercepts, then connecting them with a line. To find the - and -intercepts, we will need to substitute for one variable, solve, then repeat for the other variable.

### Finding the -intercept

Think of the point where the graph of an equation crosses the -axis. The -value of that coordinate pair is equal to and the -value is the -intercept. To find the -intercept of the given equation, we should substitute for and solve for
An -intercept of means that the graph passes through the -axis at the point

### Finding the -intercept

Let's use the same concept to find the -intercept. Consider the point where the graph of the equation crosses the -axis. The -value of the coordinate pair at the -intercept is Therefore, substituting for will give us the -intercept.
A -intercept of means that the graph passes through the -axis at the point

### Graphing the equation

We can now graph the equation by plotting the intercepts and connecting them with a line.

b
Let's start by identifying the intercepts.

### Finding the -intercept

To find the -intercept we'll substitute  for  and then solve for
An -intercept of means that the graph passes through the -axis at the point

### Finding the -intercept

To find the -intercept we do the opposite, substituting  for
A -intercept of means that the graph passes through the -axis at the point

### Graphing the equation

We can now graph the equation by plotting the intercepts and connecting them with a line.

c

### Finding the -intercept

Here, is substituted with  and then solve the equation for
The -intercept is

### Finding the -intercept

For the -intercept we'll substitute  with instead.
The -intercept is

### Graphing the equation

We can now graph the equation by plotting the intercepts and connecting them with a line.