Lesson 13

In this warm-up, students practice applying the distributive property to write equivalent quadratic expressions in standard and factored forms. They also notice that when a quadratic expression has no constant term (that is, is in the form of \(x^2 + bx\)), its factors are a variable and a sum (or a difference). Their awareness of this structure (MP7) prepares them to think about how the linear term in a quadratic expression affects the behavior of the graph later in the lesson.

Arrange students in groups of 2. Give them a moment of quiet think time, and then time to briefly share their responses with their partner.

Consider displaying the completed table for all to see. If needed, discuss only the last few expressions that can be written in a few different but equivalent ways.

Then, focus the discussion on the second question. Ask students:

• “Look at all the quadratic expressions in standard form. What do they have in common?” (They have a squared term and a linear term. They do not have a constant term.)
• “From these expressions, what can we predict, if anything, about the features of the graphs that represent them?” (Their \(y\)-intercept is \((0,0)\).) “Can you tell where the \(x\)-intercepts or the vertex will be?” (Not easily.)
• “Look at all the expressions in factored form. What do they have in common?” (They are all in the form of \(x\) or \(\text-x\) times a sum or a difference.)
• “From these expressions, what can we predict, if anything, about the features of the graphs?” (We can “see” their \(x\)-intercepts and the vertex.)

Remind students that they have looked at how the constant term and the coefficient of the squared term affect the features of a graph representing a quadratic function. Tell them that they will now explore how the linear term influences the graph, and that along the way, it can be helpful to think flexibly about the expressions that represent a function.

In this activity, students use graphing technology to experiment with the linear term in quadratic expressions of the form \(ax^2 + bx\), where \(a\) is either positive and negative. They notice that adding a linear term to \(x^2\) or \(\text-x^2\) shifts the graph both horizontally and vertically.

One way to quantify the shift of a graph is in terms of the \(x\)-intercepts and the vertex. For example, the graph of \(y=x^2\) has an \(x\)-intercept at \((0,0)\). Adding \(20x\) to \(x^2\) shifts the graph to the left and down such that the \(x\)-intercepts are now at \((\text-20,0)\) and \((0,0)\). The shift also moves the vertex of the graph from \((0,0)\) to (\(\text-10, \text-100)\). Students make sense of the values of these \(x\)-intercepts by recognizing that \(x^2+20x\) is equivalent to \(x(x+20)\), and that \(x\) values of 0 and -20 give the expression a value of 0.

Students then use their observations to predict the \(x\)-intercepts and vertex of the graphs of several equations without graphing. They also notice that when a quadratic expression has no linear term but has a constant term, rewriting it in factored form gets tricky, and that the graph may or may not have \(x\)-intercepts. (For now they are just making this observation. In a later unit, they will explore why this is the case.)

As they change the parameters of an expression on the graph, study the effects on the graph, and articulate their observations, students practice looking for and describing regularity through repeated reasoning (MP8).

Provide access to devices that can run Desmos or other graphing technology. Consider arranging students in groups of 2. For the first question, ask one partner to operate the graphing technology and the other to record the group’s observations, and then to switch roles halfway. If using Desmos, instruct students that creating a slider to experiment with linear terms might be a useful tool in this activity.

Invite students to share their observations and predictions.

To help students make sense of the shifts in \(x\)-intercepts, relate the expressions in standard form to equivalent expressions in factored form. Remind students that the factored form allows us to see the zeros of a quadratic function and the \(x\)-intercepts of the graph representing the function. Writing the expression \(x^2-20x\) as \(x(x-20)\) lets us see that the zeros are 0 and 20 (because \(0(0-20)\) and \(20(20-20)\) both give an output of 0) and the \(x\)-intercepts are \((0,0)\) and \((20,0)\). Likewise, \(x^2 + 50x\) can be written as \(x(x+50)\), so the \(x\)-intercepts of \((\text-50,0)\) and \((0,0)\) make sense.

Discuss with students:

• “What if the squared term has a negative coefficient, as in \(y=\text-x^2 -36x\)? How can we use the equivalent expression in factored form to make sense of the \(x\)-intercepts?” (Factoring out \(\text-x\) in \(\text-x^2 -36x\) gives \(\text-x(x+36)\), so the \(x\)-intercepts are at \(x=0\) and \(x=\text-36\), because \(0(0+36)\) and \(\text{-36}(\text-36+36)\) both give an output of 0.)
• “Can we still predict the \(x\)-coordinate of the vertex of a graph if the coefficient of the squared term is negative? Why or why not?” (Yes. When there are two \(x\)-intercepts, the vertex is still halfway between the two \(x\)-intercepts.)
• “What if there isn’t a linear term but there is a constant term? How did you find the \(x\)-intercepts without graphing? Were you able to write the expression in factored form?” (It was not as easy to tell what the \(x\)-intercepts were or to write the expressions in factored form. For \(y=x^2+16\), there were no \(x\)-intercepts, but for \(y=x^2-25\), there were two.)

Tell students they will continue to make sense of quadratic equations and their graphs in the rest of this unit and an upcoming unit.

Time permitting, consider asking students to predict how the graph of \(y=x^2\) would change if we add both a linear term and a constant term, for example, adding \(\text-9x\) and then \(8\). “How would the the graph of \(y=x^2 - 9x + 8\) differ from that of \(y=x^2\)?” Students are likely to say that the graph would have a \(y\)-intercept of \((0,8)\), but the \(x\)-intercepts are now harder to determine. Tell students that they will learn how to find the zeros and the \(x\)-intercepts of the graph representing such functions in a later unit.

Earlier, students were given quadratic equations and asked to reason about the graphs that represent them. Here, they reason the other way around: they write equations that can be represented by given graphs. To do so, students need to apply their insights about the structure of quadratic expressions in different forms and their connections to the graphs (MP7).

To help students write and check their equations, at least 3 points on each graph are labeled with their coordinates. In some cases, it is easier to write an equation in factored form. In other cases, an equation in standard form can be more easily derived. Identify students who are attentive to the information given in the graph and use it strategically. Invite them to share during discussion.

Provide access to graphing technology.

The activity allows for instant feedback to students (if an equation is incorrect, the graph that results will not match the given graph). Because the equations increase in complexity as students move from Graph A to Graph J, students may find that their equations are off but not know what the issue is or how to correct them. Consider arranging students in groups of 2 so that they can discuss how to write the more-challenging equations.

Select students to share their strategies for writing equations. Highlight explanations such as:

• When the graph opens up, the \(x^2\) term in standard form has a positive coefficient. When it opens down, the \(x^2\) term has a negative coefficient.
• When two \(x\)-intercepts are known, we can write the equations in factored form.
• When the \(y\)-intercept is not at \((0,0)\) and the graph is centered on the \(y\)-axis, the equation in standard form will have the form of \(ax^2+c\) (no linear term) and \((0,c)\) is the \(y\)-intercept. (A linear term would have shifted the graph horizontally as well as vertically so it would not be centered on the \(y\)-axis.)
• When \((0,0)\) is the only \(x\)-intercept, the \(y\)-intercept, and the vertex, the equation in standard form has only the squared term (\(ax^2\)). If there was a linear term, the graph would not be centered on the \(y\)-axis. If there was a constant term, the graph will move up or down and either cross the \(x\)-axis in two places or not cross it at all (that is, there would be two \(x\)-intercepts or none).

Consider discussing Graph F, as it is unlike most graphs students have seen so far (it has only one \(x\)-intercept at \((2,0)\) and one \(y\)-intercept that is not at the origin). Here a couple of approaches:

• Help students see that the factored form is still useful here. Students may guess that \((x-2)\) is one of the factors (because multiplying any number by \(2-2\) would give an output of 0) but may be unsure about the second factor. Point out that the second factor is also \((x-2)\) because there is no other \(x\)-intercept. Expanding \((x-2)(x-2)\) into standard form gives \(x^2-4x+4\), which allows us to see that the \(y\)-intercept is \((0,4)\).
• If we shift this graph down 4 units, it will intercept the \(x\)-axis at 0 and 4 and its equation would be \(y=x(x-4)\) or \(y=x^2-4x\) (per what we learned earlier). To arrive at the Graph F, we would have to move it back up 4, making the equation \(x^2-4x+4\).