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make a table of values for the following equation. then graph the equat…

Question

make a table of values for the following equation. then graph the equation.\\( y = |x + 1| \\)\\( \text{(simplify your answers.)} \\)\\( \text{graph the equation } y = |x + 1| \\). choose the correct graph below.\\( \text{help me solve this} \\)\\( \text{view an example} \\)\\( \text{get more help} \\)\\( \text{clear all} \\)\\( \text{final check} \\)

Explanation:

Step1: Analyze the absolute - value function

The general form of an absolute - value function is \(y = |x - h|+k\), where \((h,k)\) is the vertex of the V - shaped graph. For the function \(y=|x + 1|\), we can rewrite it as \(y=|x-(- 1)|+0\). So the vertex of the graph is at \((-1,0)\).

Step2: Analyze the slope of the two parts of the absolute - value function

For an absolute - value function \(y = |x - h|\), when \(x\geq h\), \(y=x - h\) (the slope is \(1\)), and when \(x\lt h\), \(y=-(x - h)=-x + h\) (the slope is \(- 1\)). For \(y = |x + 1|\), when \(x\geq - 1\), \(y=x + 1\) (slope \(m = 1\)), and when \(x\lt - 1\), \(y=-(x + 1)=-x - 1\) (slope \(m=-1\)).

Step3: Analyze the graphs

  • Let's check the vertex first. The vertex of \(y = |x + 1|\) is at \((-1,0)\).
  • For the part where \(x\geq - 1\), the slope is \(1\), so the line should be increasing with a slope of \(1\) for \(x\geq - 1\). For the part where \(x\lt - 1\), the slope is \(-1\), so the line should be decreasing with a slope of \(-1\) for \(x\lt - 1\).

Now let's analyze the options:

  • Option A: Let's check the vertex and the slopes. The vertex seems to be at \((0,1)\) (not \((-1,0)\)), so it's incorrect.
  • Option B: The vertex is at \((0,1)\) (not \((-1,0)\)), so it's incorrect.
  • Option C: The vertex is at \((-1,0)\)? No, the vertex here seems to be at \((0, - 1)\) (incorrect).
  • Option D: The vertex is at \((-1,0)\). For \(x\geq - 1\), the line has a slope of \(1\) (increasing), and for \(x\lt - 1\), the line has a slope of \(-1\) (decreasing). This matches the properties of the function \(y = |x + 1|\).

Answer:

D