Question

an absolute value function is given. f(x)=|x| graph the function that represents the transformation f(x)+3.

Explanation:

Step1: Analyze the transformation

The original function is \( f(x)=|x| \), and the transformed function is \( f(x)+3 = |x| + 3 \). This is a vertical shift transformation. For a function \( y = f(x) \), the transformation \( y=f(x)+k \) (where \( k>0 \)) shifts the graph of \( f(x) \) upward by \( k \) units. Here, \( k = 3 \), so we need to shift the graph of \( y = |x| \) upward by 3 units.

Step2: Recall the graph of \( y = |x| \)

The graph of \( y=|x| \) is a V - shaped graph with its vertex at the origin \((0,0)\), and it has two linear parts: \( y = x \) for \( x\geq0 \) and \( y=-x \) for \( x < 0 \).

Step3: Determine the vertex of the transformed graph

When we shift the graph of \( y = |x| \) upward by 3 units, the vertex of the V - shape, which was at \((0,0)\) for \( y = |x| \), will now be at \((0,0 + 3)=(0,3)\).

Step4: Determine other key points

• For the part \( y=x\) (when \( x\geq0 \)) of the original function \( y = |x| \), let's take a point, say \( x = 1 \), then \( y=|1| = 1 \). After the transformation \( y=|x|+3 \), when \( x = 1 \), \( y=|1|+3=4 \), so the point is \((1,4)\). When \( x = 2 \), \( y=|2|+3 = 5 \), so the point is \((2,5)\).
• For the part \( y=-x\) (when \( x<0 \)) of the original function \( y = |x| \), let's take \( x=-1 \), then \( y=|-1| = 1 \). After the transformation, when \( x=-1 \), \( y=|-1|+3 = 4 \), so the point is \((-1,4)\). When \( x=-2 \), \( y=|-2|+3=5 \), so the point is \((-2,5)\).

To graph \( y = |x|+3 \), we plot the vertex \((0,3)\) and the other points we found (like \((1,4)\), \((2,5)\), \((-1,4)\), \((-2,5)\)) and draw the V - shaped graph passing through these points, with the left - hand side (for \( x < 0 \)) having a slope of - 1 and the right - hand side (for \( x\geq0 \)) having a slope of 1, and the vertex at \((0,3)\).

Answer:

The graph of \( y = |x|+3 \) is a V - shaped graph with vertex at \((0,3)\), passing through points like \((1,4)\), \((2,5)\), \((-1,4)\), \((-2,5)\) (and other corresponding points), with the left branch having a slope of - 1 and the right branch having a slope of 1.