Question

graph this function:
y = 3|x|
click to plot the vertex first.

Explanation:

Step1: Identify the vertex of the absolute - value function

The general form of an absolute - value function is \(y = a|x - h|+k\), where \((h,k)\) is the vertex. For the function \(y = 3|x|\), we can rewrite it as \(y=3|x - 0|+0\). So, the vertex \((h,k)\) is \((0,0)\). We first plot the point \((0,0)\) on the coordinate plane.

Step2: Find other points to graph the function

We can choose some values of \(x\) and find the corresponding values of \(y\).

• When \(x = 1\), \(y=3|1|=3\times1 = 3\). So, the point is \((1,3)\).
• When \(x=- 1\), \(y = 3|-1|=3\times1=3\). So, the point is \((-1,3)\).
• When \(x = 2\), \(y=3|2| = 6\). So, the point is \((2,6)\).
• When \(x=-2\), \(y=3|-2|=6\). So, the point is \((-2,6)\).

After plotting the vertex \((0,0)\) and other points like \((1,3)\), \((-1,3)\), \((2,6)\), \((-2,6)\) etc., we can draw two rays. One ray passes through \((0,0)\) and \((1,3)\) (and other points with \(x\geq0\)) and the other ray passes through \((0,0)\) and \((-1,3)\) (and other points with \(x < 0\)). The graph of \(y = 3|x|\) is a V - shaped graph with the vertex at the origin \((0,0)\) and it opens upwards (since \(a = 3>0\)) and is steeper than the graph of \(y=|x|\) (because \(|a|=3>1\)).

Answer:

The vertex of the function \(y = 3|x|\) is at \((0,0)\). To graph the function, first plot the point \((0,0)\). Then plot other points such as \((1,3)\), \((-1,3)\), \((2,6)\), \((-2,6)\) etc. and draw the two rays of the V - shaped graph passing through these points with the vertex at \((0,0)\).