1.
A picture is shown of the graph of [tex]y=|x|[/tex]. To graph [tex]y=|x|-3[/tex], we use the translation property of graphs. If [tex]y=f(x)[/tex] is a function given, then [tex]y=f(x)+c[/tex] (c being a constant) will shift the curve c units up. Similarly if c is negative, it is going to shift the curve downwards c units. Hence, the graph of the function [tex]y=|x|-3[/tex] is [tex]y=|x|[/tex] shifted 3 units down. Attached picture shows the new graph.
2.
Attached picture shows the parent function [tex]f(x)=|x|[/tex] in blue and the new transformed function [tex]g(x)=-\frac{3}{4} |x|[/tex] in red.
- The negative sign in front of [tex]|x|[/tex] reflects the line about the x axis.
- The [tex]\frac{3}{4}[/tex] in front of the function makes the corresponding y values for all x less ([tex]\frac{3}{4}[/tex] of all values in the original).
3.
For the given function [tex]y=4-|x+2|[/tex], we write a general form of the absolute value function and identify the constants.
General form,
[tex]y=a|x+[tex](-b,c)[/tex]b|+c[/tex]
- If a is positive, graph opens upward and opens downward when a is negative.
- +b translates the original graph ([tex]y=|x|[/tex]) b units left and -b translates the original b units right
- +c translates the original graph c units up and -c translates it c units down
- Axis of symmetry is given as [tex]x=-b[/tex]
- Vertex is given as (-b,c)
As noted, we can now say that the vertex is [tex](-2,4)[/tex] and axis of symmetry as [tex]x=-2[/tex].
As for transformations (comparing with parent function), we can note the following:
- reflected about x axis, opens downward
- translated 2 units left
- translated 4 units up
4.
From the vertex shown be know that this graph is translated 2 units left and 6 units down. So by using the properties noted above, we can write [tex]y=|x+2|-6[/tex]. But the graph, compared to the parent function [tex]|x|[/tex] ) is a little wider. So there is a value for a, the coefficient before the absolute value sign. To figure it out, we can take any 2 points on the graph and solve for a. Let us take a random point [tex](0,-5)[/tex].
[tex]y=a|x+2|-6[/tex], pluggin 0 in x and -5 in y gives us,
[tex]-5=a|0+2|-6\\-5=2a-6\\1=2a\\a=\frac{1}{2}[/tex].
So our final equation is [tex]y=\frac{1}{2}|x+2|-6[/tex]
5.
From the equation we can make out that the graph is shifted 2 units left and 3 units down compared to the parent [tex]|x|[/tex]. Also, the 2 in front tells us all y values are twice of the original y values of the parent function for all values of x. Attached graph is shown.
