Absolute Value Lesson 3
Solving absolute value equations graphically and algebraically.
Our goal today is to take what we have seen using the Dynagraphs and look at solving absolute value problems from both a graphical approach on the coordinate plane and algebraically.
To solve |x-5|=2 graphically, we can think of it as a system of equations where Y1=|x-5| and Y2=2. How is a solution to a system of equations represented graphically? Students should recall that the solution is where the two functions intersect on the graph. Additionally students should be encouraged to connect how this approach connects to getting solutions from the Dynagraph.
Have students graph each function by hand and identify where the solutions are located on the graph. [x=3 and x=7.]
How can we find the solutions algebraically?
If we think back to Lesson 2, we can write the absolute value function as a piecewise function consisting of two lines:
x-5 for the domain values where the output y=x-5 is above the x-axis
and the opposite of x-5 for the domain values where the output y=x-5 is below the x-axis.
We are then figuring out where each of these pieces of the function are equal to 2 to solve the equation.
There is a difference in the way we are setting up our equations for the piecewise function and for solving the absolute value problem. Did you notice we did not take the function x-5 and set it equal to 2 and -2? Instead we took the function equal to 2 and the opposite of the function equal to 2.
We are thinking about -(x-5) as being a reflection over the x-axis for values in the domain that had a negative output value. So, instead of thinking of where that portion of the graph is equal to -2 we reflect the function (or take the opposite of it) and figure out where it is equal to 2. We are finding values for the function x-5 that are a distance of 2 away from the x-axis.
Have students complete the Solving Absolute Value Equations AS. [SMP 5, SMP 7]
Discuss student work from Solving Absolute Value Equations activity sheet. Make sure to address why |x - 6|=-4 has no solution.
If time permits, consider the following extension.
After discussing the first example, have students consider another absolute value equation: |x^2-9x+14|=3. Ask students, "What is a solution? What is a non-solution? Explain."
Students will most likely struggle to find a solution to this equation because the solutions turn out to be irrational. This might be a good time to discuss the pros and cons of solving absolute value equations by graphing as compared to algebraically.
Have students graph the system: Y1=|x^2-9x+14| and Y2=3. (Recall, from lesson 2, that you can graph 〖|x〗^2-9x+14| by graphing the quadratic x^2-9x+14 as long as you reflect the portion that has negative output values over the x-axis.)
Ask students to rewrite the absolute value equation using a piecewise function. The portion of the parabola that is below the x-axis is reflected over the x-axis since it is an absolute value; that is why we can write that part of the function as the "opposite" or -(x^2+2x-3).
So, solving this system algebraically, students can set each of these pieces equal to 3 and solve for x.
Finally, have students consider the following question: Why is there no solution to the following problem: |x〖^2〗-4|=-2? [SMP 7]
[The students may say that there is no value for x that makes the equation true. If needed, help students also notice that the system of equations to be graphed will not have an intersection.]
Summarize or explain how to rewrite an absolute value function as a piecewise function.
How does solving absolute value equations graphically as a system of equations support the process used to solve absolute value equations algebraically? When is one method more appropriate than the other?
Leave your thoughts in the comments below.
CCSS, Content Standards to specific grade/standard
CCSS, Standards for Mathematical Practices
PtA, highlighted Effective Teaching Practice and/or Guiding Principle CCSS