Core Connections Algebra 2, 2013
CC
Core Connections Algebra 2, 2013 View details
1. Section 4.1
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Exercise 23 Page 176

Write the equation as a system of two equations. Graph the two equations by hand to find the points of intersection. The absolute value function can be graphed using a series of transformations. The second function is graphed using its slope-intercept form.

Solutions: x=3, x=6
Diagram:

Practice makes perfect

To solve this equation graphically, we will treat each side as a separate function and form a system of equations. 2|x-4|-3= 23x-3 ⇓ f(x)= 2|x-4|-3 [0.3em] g(x) = 23x-3 The solutions to the system are the x-values that satisfy both equations. We can find these graphically by plotting both functions and identifying the x-values where they intersect.

Graphing the Functions

To graph f(x), we should consider the general equation of an absolute value function.

y= a|x-h|+k [-1.2em] Horizontal Translation:& h Vertical Translation:& k Vertical Stretch:& a As we can see, the constants and the coefficient control different transformations of the parent function. y=|x| When we subtract from the input, we get a translation to the right. When we subtract from the output, we get a translation down. Finally, when a is greater than 1, we get a vertical stretch. |c|l| Right-Hand Side & Transformation 2|x-4|-3 & &Vertical stretch by a &factor of 2. 2|x- 4|-3 & &Horizontal translation &to the right by 4 units. 2|x- 4|-3 & &Vertical translation &down by 3 units. Let's show these transformations starting with the translations.

Next we will stretch the graph by a factor of 2. The stretching is done with respect to the horizontal line through the new locator point at (4,-3). In other words, all points on the graph moves twice the distance away from this line.

To graph the straight line, we need at least two points that fall on the line. Since the function is written in slope-intercept form, one of the points is the y-intercept. To get a second point, we will use the line's slope. When we have identified these points, we can draw the line.

Finding the Solutions

To find the solutions, we have to determine the x-coordinates of the graph's points of intersection. Examining the diagram, we see that the graph's intersect at x=3 and x=6.

The graphs intersect at x=3 and x=6. These values solves the original equation.