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The graph of a piecewise function is shown.The graph of a linear function starts in quadrant 3, has a positive slope, and a right endpoint at open circle left-parenthesis negative 2 comma 1 right parenthesis. It also shows a portion of a quadratic function opening upwards with left endpoint closed circle left-parenthesis negative 2 comma 3 right-parenthesis, vertex left-parenthesis 0 comma negative 1 right-parenthesis, and right endpoint closed circle left-parenthesis 1 comma 0 right-parenthesis. It also shows the graph of a logarithmic function with left endpoint open circle left-parenthesis 1 comma 2 right-parenthesis. The logarithmic function increases as it goes to the right and passes through the point left-parenthesis 5 comma 3 right parenthesis.
Question
Use the graph to answer the question.
A student was given the piecewise functionf left parenthesis x right parenthesis equals enlarged left brace start layout 1st row x plus 3 comma x less than or equal negative 2 2nd row x squared minus 1 comma negative 2 less than x less than 1 3rd row log subscript 2 baseline left parenthesis negative x plus 3 right parenthesis comma 1 less than or equals x less than 3 end layout and created this graph of the function.

Is the student’s graph a correct representation of the function? If not, explain how it should be corrected.
(3 points)

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Answer 1

To determine if the student's graph of the piecewise function \( f(x) \) is a correct representation, we need to analyze each piece of the function according to the defined intervals.

The piecewise function is defined as follows:

1. For \( x \leq -2 \): \( f(x) = x + 3 \)

   • This is a linear function with a slope of 1. The graph starts in quadrant 3 and, at \( x = -2 \), should yield \( f(-2) = -2 + 3 = 1 \). This is represented by an open circle at \( (-2, 1) \).

2. For \( -2 < x < 1 \): \( f(x) = x^2 - 1 \)

   • This is a quadratic function that opens upwards. At \( x = -2 \), \( f(-2) = (-2)^2 - 1 = 4 - 1 = 3 \), which corresponds to a closed circle at \( (-2, 3) \). The vertex is located at \( (0, -1) \) and the value at \( x = 1 \) is \( f(1) = 1^2 - 1 = 0 \), which corresponds to a closed circle at \( (1, 0) \).

3. For \( 1 \leq x < 3 \): \( f(x) = \log_2(-x + 3) \)

   • The logarithmic function will be evaluated starting at \( x = 1 \), where \( f(1) = \log_2(-1 + 3) = \log_2(2) = 1 \). This part of the function has an open circle at \( (1, 2) \) and increases, passing through the point \( (5, 3) \), which is consistent with logarithmic behavior.

Analysis of the Student's Graph

• The linear part should start in quadrant 3 and show an open circle at \( (-2, 1) \). It should have a 1 in the y-coordinate at \( (-2, 1) \) and extend downward to the left.
• The quadratic part should start at \( (-2, 3) \) (closed circle) and go through the vertex at \( (0, -1) \) and end at \( (1, 0) \) (closed circle).
• The logarithmic part should start at \( (1, 2) \) (open circle), and then it should increase, passing through \( (5, 3) \).

Corrections Needed

1. The graph of the linear function needs to be adjusted to reflect that it ends with an open circle at \( (-2, 1) \), not a closed circle.
2. The quadratic part should correctly start from a closed circle at \( (-2, 3) \), pass through the vertex, and end at \( (1, 0) \) with a closed circle.
3. Ensure the logarithmic part starts at \( (1, 2) \) (open circle) and behaves as described above.

If these characteristics are correctly updated in the student’s graph, it will accurately represent the given piecewise function \( f(x) \).