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Calculus IChapter 1: ReviewSection 1-10

Common Graphs — Question 2

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Question 2

Sketch the graph of the function: f(x)=|x+1|−2f(x) = |x + 1| - 2

Instructions: Identify the vertex, axis of symmetry, domain, range, and x- and y-intercepts. Then sketch the graph.

Original worksheet page 1: question and worked solution for 1-10-002
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Question 2 - Solution

We are given: f(x)=|x+1|−2f(x) = |x + 1| - 2

Step 1: Transform the parent function f(x)=|x|f(x) = |x|

  • Shift left by 1: x→x+1x \to x + 1

  • Shift down by 2: f(x)→f(x)−2f(x) \to f(x) - 2

Vertex: The vertex of f(x)=|x|f(x) = |x| is at (0,0)(0, 0), so for this function it’s at (−1,−2)(-1, -2)

Axis of Symmetry: x=−1x = -1

Domain: (−∞,∞)(-\infty, \infty)

Range: [−2,∞)[-2, \infty)

X-intercepts: |x+1|−2=0⇒|x+1|=2⇒x+1=±2⇒x=1,−3|x + 1| - 2 = 0 \Rightarrow |x + 1| = 2 \Rightarrow x + 1 = \pm 2 \Rightarrow x = 1, -3

Y-intercept: f(0)=|0+1|−2=1−2=−1f(0) = |0 + 1| - 2 = 1 - 2 = -1

See the diagram in the original worksheet below.

Original worksheet page 2: question and worked solution for 1-10-002

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