Lesson 11

This warm-up refreshes the work in an earlier lesson. It prepares students to deepen their understanding of the factored form and the intercepts of a graph that represents a quadratic function.

As students work, notice how they find the \(y\)-coordinate of the \(y\)-intercept. Identify students who do so by evaluating \(w(0)\). Ask them to share during whole-class discussion.

Arrange students in groups of 2. Give students quiet work time and then time to share their work with a partner. Follow with a whole-class discussion.

Some students may think that the numerical values in the equation correspond directly to the \(x\)-intercepts in the graph and incorrectly state that \(a= \text-2\) and \(c=1.6\). Remind them that a graph shows all pairs of \(x\) and \(y\) values that make the equation true. Consider asking these students to try substituting -2 for \(x\) and evaluating the expression to verify that \(w(\text-2)=0\).

Students will have opportunities to attend to the signs or the operations in quadratic expressions in factored form, so it is not essential that this misconception is corrected at this moment.

Invite students to share their responses and reasoning. If not mentioned in students’ explanations, emphasize that:

• \(b\) and \(d\) must be 0 because the \(y\)-coordinate of the \(x\)-intercepts of the graph of any function is 0.
• \(e\) must be 0 because the \(x\)-coordinate of the \(y\)-intercept of the graph of any function is 0.
• \(a\) and \(c\) correspond to the “1.6” and “2” in the factored expression. Students may reason that the graph shows that the positive \(x\)-intercept is farther away from 0 than the negative \(x\)-intercept, suggesting that \(c\) is 2 and \(a\) is -1.6. They may also use their observations from an earlier lesson: that when a factor in the expression shows a negative sign or subtraction, a corresponding intercept takes a positive value, and when a factor shows a positive sign or addition, a corresponding intercept takes a negative value. Either explanation is reasonable at this point.
• \((0,f)\) is the \(y\)-intercept, so the value of \(f\) be found by evaluating \(w(0)\), which is -3.2, because \((0+1.6)(0-2) = (1.6)(\text-2) = \text-3.2\).

In this activity, students continue to make sense of their earlier observations about the connection between the factored form of a quadratic expression and the \(x\)-intercepts of its graph. They study two very similar expressions: \(x(x+4)\) and \(x(x-4)\). They evaluate each at different \(x\) values. The expressions are chosen so that students can shift their focus from the numbers to the operations, attending to precision (MP6) along the way.

Students notice that for \(x(x+4)\), the \(x\) values that produce a 0 output are 0 and -4, and for \(x(x-4)\), those values are 0 and 4. This helps to explain why the signs of the \(x\)-intercepts are the opposite of the signs in the factors. In an upcoming unit, they will further develop this understanding algebraically, in terms of the zero product property. For now, it suffices that they verify the \(x\)-intercepts by evaluating the quadratic expression at the predicted values and checking that the outputs are 0.

Students also observe, by looking for regularity in repeated reasoning, that the horizontal location of the vertex of the graph can be identified once the \(x\)-intercepts are known: it is exactly halfway between the intercepts (MP8). This suggests that a quadratic expression in factored form can help us “see” both the \(x\)-coordinates of the \(x\)-intercepts and the \(x\)-coordinate of the vertex of the graph. Some students may notice the horizontal location of the vertex could be determined using the halfway point between any two points with the same \(y\)-coordinate because the graphs have reflection symmetry across a vertical line through the vertex.

As students work, look for those who use the table to determine the \(x\)-intercepts, rather than by plotting the points first. Invite those students to share their strategy during the synthesis.

In this activity, students are generalizing the relationship between the zeros on the graph of a function and the factored form of the function so technology is not an appropriate tool.

Display for all to see the equations defining \(f\) and \(g\). 

\(f(x) = x(x+4)\) and \(g(x) = x(x-4)\)

Poll to class to gather some predictions about the graphs that represent the two functions and display the results for all to see:

• In what ways would the graphs be alike?
• In what ways would they be different?

Arrange students in groups of 2. Ask partners to split the work on the first question (with one person analyzing function \(f\) and the other person analyzing function \(g\)) and then working together to plot the points and make observations in the last question. 

If needed, remind students that the vertex of a graph is the point at which the graph changes direction from increasing to decreasing, or from decreasing to increasing.

Since this activity was designed to be completed without technology, ask students to put away any devices.

If desired, distribute colored pencils so that each student has access to two different colors for creating their graphs in the last question.

Discuss how students use the completed tables to help them find the \(x\)-intercepts and the vertex of each graph. If not already mentioned in students’ explanations, highlight that:

• The \(x\) values that produce 0 for the output are the zeros of the function. They tell us where the graph intersects the \(x\)-axis. For \(f(x) = x(x+4)\), those values are 0 and -4, so the \(x\)-intercepts are \((0,0)\) and \((\text-4,0)\). For \(g(x) =x(x-4)\), those values are 0 and 4, so the \(x\)-intercepts are \((0,0)\) and \((4,0)\).
• The vertex tells us the point where the function reaches its maximum or minimum value. We can see the minimum value in each table. In the first table, \(f(\text-2)\) is the lowest value. Because the values on one side of \(f(\text-2)\) mirror those on the other, we can reason that -4 is the lowest value of the function \(f\). In the second table, \(g(2)\) is the lowest value.
• The vertex is halfway between each pair of \(x\)-intercepts.

If time permits, consider asking students how they think the graphs of the functions given by \(p(x) = (x-1)(x-3)\) and \(q(x) = (x+1)(x+3)\) will compare.

• “Where will their \(x\)-intercepts be?” (\((1,0)\) and \((3,0)\) for the graph of function \(p\), and \((\text-1,0)\) and \((\text-3,0)\) for the graph of function \(q\))
• “Where will the \(x\)-coordinate of the vertex of each graph be?” (2 for the graph of \(p\) and -2 for that of \(q\))
• “How do we find the \(y\)-coordinate of the vertex?” (Evaluate \(p(2)\) and \(q(\text-2)\).)
• “The \(y\)-intercept of the both graphs is \((0,3)\). What is another point on each graph with the same \(y\)-coordinate?” (The point 2 units to the right of the vertex of \(p\) will have same \(y\)-coordinate, so \((4,3)\). The point \((\text-4,3)\) is on the graph of \(q\).)

Clarify that a table of values won’t always show the maximum or minimum values of a function. It also won’t always help us identify the \(x\)-intercepts of a graph or the zeros of a quadratic function, especially if the zeros are not integers or the given expressions are more complex. Students will learn other ways to find the zeros of any function in another unit later in the course.

In this activity, students apply what they learned in earlier lessons to identify the \(x\)-intercepts and the vertex of the graphs representing several quadratic functions. For example, they saw that the \(x\)-coordinate of the vertex is always halfway between those of the \(x\)-intercepts, and the coordinate pairs on one side of the vertex mirror those on the other side. These observations help students locate the vertex of a graph once the \(x\)-intercepts are known, and ultimately to sketch a graph of a quadratic function using at least three identifiable points. 

The given expressions here are in factored form, but some are unlike what students have previously seen, so students will need to transfer and generalize the reasoning strategies from earlier work. They also use graphing technology to graph the functions and check their predictions.

In a later lesson, students will use the symmetry of the graph of a quadratic function to sketch graphs when they know the vertex and one additional point rather than the vertex and \(x\)-intercepts.

Focus the discussion on how students determined the \(x\)-intercepts and the \(x\)-coordinate of the vertex of a graph, and how the coordinates of these points could help them sketch the graph. Ask questions such as:

• “For \(g(x) = 2x(x-3)\), how did you find the \(x\)-intercepts without graphing?” (The \((x-3)\) suggests that one \(x\)-intercept is \((3,0)\), and evaluating \(g(3)\) does give an output of 0. The factor \(2x\) suggests that the second \(x\)-intercept is \((0,0)\), because \(2(0)\) is 0, and multiplying 0 by any number gives 0.)
• “How did you find the \(x\)-value of the vertex?” (By finding the halfway point between 0 and 3, which is 1.5.)
• “How would you find the \(y\)-coordinate of the vertex?” (By evaluating \(g(1.5)\), which gives -4.5.)
• “The expression that defines function \(h\) has the factor \((4-x)\), where the constant term appears first and \(x\) is subtracted from it. Did this affect how you determined the \(x\)-intercepts? How so?”
• “How did you sketch the graph representing \(y=(x-7)(x+11)\)?” (By finding the \(x\)-intercepts and the vertex. The intercepts are \((7,0)\) and \((\text-11,0)\), so the \(x\)-coordinate of the vertex is -2, and the \(y\)-coordinate is \((\text-2-7)(\text-2+11)\) or -81. Those points are enough to sketch a graph.)