Week 12 – Absolute Value Functions

This week, I have learned how to graph the absolute value of a linear function, the absolute value of a quadratic function and how to write it in piecewise notation.

Example of a linear function:

y = I3x-5I

Let’s start by graphing this.

We know the y-intercept is -5, and the slope is 3. Start by plotting the given information on the graph for y = 3x-5

Now, to be able to place the absolute value, we must remember that absolute values cannot be negative. So the line that goes past the x-axis will instead change to positive values which will make the line bounce upwards after touching the x-axis.

To write this in piecewise notation, we can’t really tell what the x-axis is here, so we can use the equation to determine the x-intercept.

 

 

Let’s look at an example of a quadratic function:

 

Let’s start by graphing this. We know the vertex is (-1, 1) and it’s congruent to y=x^2 which follows the pattern 1,3,5,7,9…

Once you have it graphed without the absolute value symbol, you now can start graphing the equation when it’s in the absolute symbol. To do this, all the negative values must switch to positive values. Therefore, the outsides of the parabola will flip up.


Now that you have this, we can use this graph to determine the intercepts and domain and range. The x-intercept is -2 and 0 and the y-intercept is 0. The domain is always the element of all real numbers (xER) and the range is y ≥ 0 as the values of y must be +.

To write this in piecewise notation, we must first look at the x-intercepts. We can see that the outer part changes but the inner part does not whether it’s in the absolute value symbol or not. So y = -(x+1)^2+1 is if  -2 ≥ x ≥ 0 (The inner part of the graph; we are including the + values and 0).

For y=-(-(x+1)^2+1) if x < -2 and x > 0 (The outer parts of the parabola; all the negative values).

So the piecewise notation is: