QUESTION IMAGE
Question
- $f(x)=vert x + 2vert + 1$
Step1: Identify the parent function and transformations
The parent function is \( y = |x| \), which has a vertex at \( (0,0) \), opens upwards. The given function is \( f(x)=|x + 2|+1 \). For absolute - value functions of the form \( y=|x - h|+k \), the vertex is at \( (h,k) \). Here, \( h=- 2 \) and \( k = 1 \), so the vertex of \( f(x) \) is at \( (-2,1) \).
Step2: Find key points
- When \( x=-2 \), \( f(-2)=| - 2 + 2|+1=0 + 1=1 \) (vertex).
- When \( x=-3 \), \( f(-3)=| - 3+2|+1=|-1| + 1=1 + 1=2 \).
- When \( x=-1 \), \( f(-1)=| - 1 + 2|+1=|1|+1=1 + 1=2 \).
- When \( x=-4 \), \( f(-4)=| - 4 + 2|+1=|-2|+1=2 + 1=3 \).
- When \( x = 0 \), \( f(0)=|0 + 2|+1=2 + 1=3 \).
Step3: Plot the points and draw the graph
Plot the points \( (-4,3) \), \( (-3,2) \), \( (-2,1) \), \( (-1,2) \), \( (0,3) \) on the coordinate plane. Then, draw two rays: one with a slope of \( 1 \) (for \( x\geq - 2 \)) passing through the points \( (-2,1) \), \( (-1,2) \), \( (0,3) \) and one with a slope of \( - 1 \) (for \( x < - 2 \)) passing through the points \( (-4,3) \), \( (-3,2) \), \( (-2,1) \).
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The graph of \( f(x)=|x + 2|+1 \) is a V - shaped graph (absolute - value graph) with vertex at \( (-2,1) \), opening upwards. The key points are \( (-4,3) \), \( (-3,2) \), \( (-2,1) \), \( (-1,2) \), \( (0,3) \) and the graph is symmetric about the line \( x=-2 \).