Unit 7 Introduction To Quadratic Functions — Unit Plan

In this lesson, we looked at the relationship between the side lengths and the area of a rectangle when the perimeter is unchanged.

To find the width given a length $\ell$, we can write: $w= 20- \ell$.

The relationship between the length and the width is linear. If we plot the points from the table representing the length and the width, they form a line.

What about the relationship between the side lengths and the area of rectangles with a perimeter of 40 inches?

Here are some possible areas of different rectangles that have a perimeter of 40 inches.

Here is a graph of the lengths and areas from the table:

Notice that, initially, as the length of the rectangle increases (for example, from 5 to 10 inches), the area also increases (from 75 to 100 square inches). Later, however, as the length increases (for example, from 12 to 15), the area decreases (from 96 to 75).

We have not studied relationships like this yet and will investigate them further in this unit.

In this lesson, we saw some quantities that change in a particular way, but the change is neither linear nor exponential. Here is a pattern of shapes, followed by a table showing the relationship between the step number and the number of small squares.

Three steps of a growing pattern. Step 1: two squares, one atop the other. Step 2: two squares on row 1, two squares on row 2 and one square on the left on row 3. Step 3: three squares on row 1, three squares on Row 2, 3 squares on Row 3 and 1 square on the left on row 4.

The number of small squares increases by 3, and then by 5, so we know that the growth is not linear. It is also not exponential because it is not changing by the same factor each time. From Step 1 to Step 2, the number of small squares grows by a factor of $\frac{5}{2}$, while from Step 2 to Step 3, it grows by a factor of 2.

From the diagram, we can see that in Step 2, there is a 2-by-2 square plus 1 small square added on top. Likewise, in Step 3, there is a 3-by-3 square with 1 small square added. We can reason that the $n$th step is an $n$-by-$n$ arrangement of small squares with an additional small square on top, giving the expression $n^2 + 1$ for the number of small squares.

The relationship between the step number and the number of small squares is a quadratic relationship, because it is given by the expression $n^2 + 1$, which is an example of a quadratic expression. We will investigate quadratic expressions in depth in future lessons.

The distance traveled by a falling object in a given amount of time is an example of a quadratic function. Galileo is said to have dropped balls of different mass from the Leaning Tower of Pisa, which is about 190 feet tall, to show that they travel the same distance in the same time. In fact the equation $d = 16t^2$ models the distance $d$, in feet, that a metal ball falls after $t$ seconds, no matter what its mass.

Because $16 \boldcdot 4^2 = 256$, and the tower is only 190 feet tall, a metal ball hits the ground before 4 seconds.

Here is a table showing how far a metal ball has fallen over the first few seconds.

time (seconds)	distance fallen (feet)
0	0
1	16
2	64
3	144

Here are the time and distance pairs plotted on a coordinate plane:

Graph of the quadratic function $y=16t^2$on a coordinate plane, origin $O$. Horizontal axis scale 0 to 6 by 1’s, labeled “time (seconds)”. Vertical axis scale 0 to 200 by  50’s, labeled “distance dropped (feet)”. The four points of the function shown are $O$(0 comma 0), (1 comma 16), (2 comma 64) and (3 comma 144).

Notice that the distance fallen is increasing each second. The average rate of change is increasing each second, which means that the metal ball is speeding up over time. This comes from the influence of gravity, which is represented by the quadratic expression $16t^2$. It is the exponent 2 in that expression that makes it increase by larger and larger amounts.

Another way to study the change in the position of the metal ball is to look at its distance from the ground as a function of time.

Here is a table showing the distance from the ground in feet at 0, 1, 2, and 3 seconds.

time (seconds)	distance from the ground (feet)
0	190
1	174
2	126
3	46

Here are those time and distance pairs plotted on a coordinate plane:

Graph of the quadratic function $y=190 - 16t^2$ on a coordinate plane, origin $O$. Horizontal axis scale 0 to 6 by 1’s, labeled “time (seconds)”. Vertical axis scale 0 to 200 by  50’s, labeled “distance from ground (feet)”. The four points of the function shown are $O$(0 comma 190), (1 comma 174), (2 comma 126) and (3 comma 46).

The expression that defines the distance from the ground as a function of time is $190 - 16t^2$. It tells us that the metal ball's distance from the ground is 190 feet before it is dropped and has decreased by $16t^2$ when $t$ seconds have passed.

In this lesson, we looked at the height of objects that are launched upward and then come back down because of gravity.

An object is thrown upward from a height of 5 feet with a velocity of 60 feet per second. Its height, $h(t)$, in feet, after $t$ seconds is modeled by the function $h(t) = 5 + 60t - 16t^2$.

• The linear expression $5 + 60t$ represents the height that the object would have at time $t$ if there were no gravity. The object would keep going up at the same speed at which it was thrown. The graph would be a line with a slope of 60, which relates to the constant speed of 60 feet per second.
• The expression $\text-16t^2$ represents the effect of gravity, which eventually causes the object to slow down, stop, and start falling back again.

Notice the graph intersects the vertical axis at 5, which means that the object was thrown into the air from 5 feet off the ground. The graph indicates that the object reaches its peak height of about 60 feet after a little less than 2 seconds. That peak is the point on the graph where the function reaches a maximum value. At that point, the curve changes direction, and the output of the function changes from increasing to decreasing. We call that point the vertex of the graph.

Here is the graph of $h$.

Graph of the quadratic function $h(t) = 5 + 60t - 16t^2$ on a coordinate plane, origin $O$. Horizontal axis scale 0 to 4 by 1’s, labeled “time (seconds)”. Vertical axis scale 0 to 80 by  20’s, labeled “distance above ground (feet)”. Some of the points of this function are (0 comma 5), (1 comma 49), to a maximum near (1 point 9 comma 61 point 2 5) then decreasing through (2 comma 61), (3 comma 41) and (3.8 comma 0).

The graph representing any quadratic function is a special kind of “U” shape called a parabola. You will learn more about the geometry of parabolas in a future course. Every parabola has a vertex, because there is a point at which it changes direction—from increasing to decreasing, or the other way around.

The object hits the ground a little before 4 seconds. That time corresponds to the horizontal intercept of the graph. An input value that produces an output of 0 is called a zero of the function. A zero of function $h$ is approximately 3.8, because $h(3.8) \approx 0$.

In this situation, input values less than 0 seconds or more than about 3.8 seconds would not be meaningful, so an appropriate domain for this function would include all values of $t$ between 0 and about 3.8.

A quadratic function can often be defined by many different but equivalent expressions. For example, we saw earlier that the predicted revenue, in thousands of dollars, from selling a downloadable movie at $x$ dollars can be expressed with $x(18-x)$, which can also be written as $18x - x^2$.

Sometimes a quadratic expression is a product of two factors that are each a linear expression, for example $(x+2)(x+3)$. We can write an equivalent expression by thinking about each factor, the $(x+2)$ and $(x+3)$, as the side lengths of a rectangle, with each side length being decomposed into a variable expression and a number.

Rectangle, divided into 4 smaller rectangles. Top side of rectangle labeled x and 2. Left side of rectangle labeled x and 3. Small rectangles labeled as follows: Top left, x squared. Top right, 2 x. Bottom right, 6. Bottom left, 3 x.

Notice that the diagram illustrates the distributive property being applied. Each term of one factor (say, the $x$ and the 2 in $x+2$) is multiplied by every term in the other factor (the $x$ and the 3 in $x+3$).

In general, when a quadratic expression is written in the form of $(x+p)(x+q)$, we can apply the distributive property to rewrite it as $x^2 + px + qx + pq$, or as $x^2 + (p+q)x + pq$.

A quadratic function can often be represented by many equivalent expressions. For example, a quadratic function, $f$, might be defined by $f(x) = x^2 + 3x + 2$. The quadratic expression $x^2 + 3x + 2$ is called the standard form, the sum of a multiple of $x^2$ and a linear expression ($3x+2$ in this case).

In general, standard form is written as $\displaystyle ax^2 + bx + c$ 

We refer to $a$ as the coefficient of the squared term $x^2$, $b$ as the coefficient of the linear term $x$, and $c$ as the constant term.

Function $f$ can also be defined by the equivalent expression $(x+2)(x+1)$. When the quadratic expression is a product of two factors where each one is a linear expression, this is called the factored form.

An expression in factored form can be rewritten in standard form by expanding it, which means multiplying out the factors. In a previous lesson we saw how to use a diagram and to apply the distributive property to multiply two linear expressions, such as $(x+3)(x+2)$. We can do the same to expand an expression with a sum and a difference, such as $(x+5)(x-2)$, or to expand an expression with two differences, for example, $(x-4)(x-1)$.

To represent $(x-4)(x-1)$ with a diagram, we can think of subtraction as adding the opposite:

	$x$	$\text-4$
$x$	$x^2$	$\text-4x$
$\text-1$	$\text-x$	$4$

Diagram showing distributive property. Row 1: x minus four times x minus 1. Row 2: equals x plus negative 4 times x plus negative 1. Two arrows drawn from both first x and from negative 4, for each, one arrow to the second x, one arrow to negative 1. Row 3: equals x times the quantity x plus negative one, plus negative 4 times the quantity x plus negative 1. 2 arrows drawn from first x to second x and negative 1. 2 arrows drawn from negative 4 to third x and negative 1. Row 4: equals x squared plus negative 1 x plus negative 4 x plus negative 4 times negative 1. Row 5: equals x squared plus negative 5 x plus 4. Row 6: equals x squared minus 5 x plus 4.

Function $f$, given by $f(x) = (x+1)(x-3)$, is written in factored form. Recall that this form is helpful for finding the zeros of the function (where the function has the value 0) and for telling us the $x$-intercepts on the graph that represents the function.

Here is a graph representing $f$. It shows two $x$-intercepts: one at $x = \text-1$ and one at $x = 3$.

If we use -1 and 3 as inputs to $f$, what are the outputs?

• $f(\text-1)=(\text-1+1)(\text-1-3)=(0)(\text-4)=0$
• $f(3)=(3+1)(3-3)=(4)(0)=0$

Because the inputs -1 and 3 produce an output of 0, they are the zeros of function $f$. And because both $x$ values have 0 for their $y$ value, they also give us the $x$-intercepts of the graph (the points where the graph crosses the $x$-axis, which always have a $y$-coordinate of 0). So, the zeros of a function have the same values as the $x$-coordinates of the $x$-intercepts of the graph of the function.

The factored form can also help us identify the vertex of the graph, which is the point where the function reaches its minimum value. Notice that due to the symmetry of the parabola, the $x$-coordinate of the vertex is 1, and that 1 is halfway between -1 and 3. Once we know the $x$-coordinate of the vertex, we can find its $y$-coordinate by evaluating the function: $f(1) = (1+1)(1-3) = 2 (\text-2) = \text-4$. So the vertex is at $(1,\text-4)$.

When a quadratic function is in standard form, the $y$-intercept is clear: its $y$-coordinate is the constant term $c$ in $ax^2 +bx+c$. To find the $y$-intercept from factored form, we can evaluate the function at $x =0$, because the $y$-intercept is the point at which the graph has an input value of 0. $f(0) = (0+1)(0-3) = (1)(\text-3) = \text-3$.

Remember that the graph representing any quadratic function is a shape called a parabola. People often say that a parabola “opens upward” when the lowest point on the graph is the vertex (where the graph changes direction), and “opens downward” when the highest point on the graph is the vertex. Each coefficient in a quadratic expression written in standard form $ax^2 + bx+ c$ tells us something important about the graph that represents it.

The graph of $y=x^2$ is a parabola opening upward with vertex at $(0,0)$. Adding a constant term 5 gives $y = x^2 + 5$ and raises the graph by 5 units. Subtracting 4 from $x^2$ gives $y=x^2-4$ and moves the graph 4 units down.

$x$	-3	-2	-1	0	1	2	3
$x^2$	9	4	1	0	1	4	9
$x^2+5$	14	9	6	5	6	9	14
$x^2-4$	5	0	-3	-4	-3	0	5

A table of values can help us see that adding 5 to $x^2$ increases all the output values of $y = x^2$ by 5, which explains why the graph moves up 5 units. Subtracting 4 from $x^2$ decreases all the output values of $y = x^2$ by 4, which explains why the graph shifts down by 4 units.

In general, the constant term of a quadratic expression in standard form influences the vertical position of the graph. An expression with no constant term (such as $x^2$ or $x^2 +9x$) means that the constant term is 0, so the $y$-intercept of the graph is on the $x$-axis. It’s not shifted up or down relative to the $x$-axis.

The coefficient of the squared term in a quadratic function also tells us something about its graph. The coefficient of the squared term in $y = x^2$ is 1. Its graph is a parabola that opens upward.

• Multiplying $x^2$ by a number greater than 1 makes the graph steeper, so the parabola is narrower than that representing $x^2$.
• Multiplying $x^2$ by a number less than 1 but greater than 0 makes the graph less steep, so the parabola is wider than that representing $x^2$.
• Multiplying $x^2$ by a number less than 0 makes the parabola open downward.

Coordinate plane, 4 graphs of quadratic functions, all with the maximum or minimum at the origin. First, y = 2 x squared, opens up. Second, y = x squared, opens up but wider than the first. Third, y = fraction 1 over 2 x squared, opens up wider than the first 2. Fourth, y = negative 2 x squared, opens down.

$x$	-3	-2	-1	0	1	2	3
$x^2$	9	4	1	0	1	4	9
$2x^2$	18	8	2	0	2	8	18
$\text-2x^2$	-18	-8	-2	0	-2	-8	-18

If we compare the output values of $2x^2$ and $\text-2x^2$, we see that they are opposites, which suggests that one graph would be a reflection of the other across the $x$-axis. 

In an earlier lesson, we saw that a quadratic function written in standard form, $ax^2 + bx +c$, can tell us some things about the graph that represents it. The coefficient $a$ can tell us whether the graph of the function opens upward or downward, and also gives us information about whether it is narrow or wide. The constant term $c$ can tell us about its vertical position.

Recall that the graph representing $y = x^2$ is an upward-opening parabola with the vertex at $(0,0)$. The vertex is also the $x$-intercept and the $y$-intercept.

Suppose we add 6 to the squared term: $y=x^2+6$. Adding a 6 shifts the graph upward, so the vertex is at $(0,6)$. The vertex is the $y$-intercept, and the graph is centered on the $y$-axis.

What can the linear term $bx$ tell us about the graph representing a quadratic function?

The linear term has a somewhat mysterious effect on the graph of a quadratic function. The graph seems to shift both horizontally and vertically. When we add $bx$ (where $b$ is not 0) to $x^2$, the graph of $y=x^2+bx$ is no longer centered on the $y$-axis.

Suppose we add $6x$ to the squared term: $y=x^2+6x$. Writing the $x^2+6x$ in factored form as $x(x+6)$ gives us the zeros of the function, 0 and -6. Adding the term $6x$ seems to shift the graph to the left and down and the $x$-intercepts are now $(\text-6,0)$ and $(0,0)$. The vertex is no longer the $y$-intercept, and the graph is no longer centered on the $y$-axis.

What if we add $\text-6x$ to $x^2$? We know that $x^2-6x$ can be rewritten as $x(x-6)$, which tells us the zeros: 0 and 6. Adding a negative linear term to a squared term seems to shift the graph to the right and down. The $x$-intercepts are now $(0,0)$ and $(6,0)$. The vertex is no longer the $y$-intercept, and the graph is not centered on the $y$-axis.

Sometimes the expressions that define quadratic functions are written in vertex form. The function $f(x) = (x-3)^2+4$ is in vertex form and is shown in this graph.

The vertex form can tell us about the coordinates of the vertex of the graph of a quadratic function. The expression $(x-3)^2$ reveals that the $x$-coordinate of the vertex is 3, and the constant term, 4, reveals that the $y$-coordinate of the vertex is 4. Here the vertex represents the minimum value of function $f$, and its graph opens upward.

In general, a quadratic function expressed in vertex form is written as $\displaystyle y = a(x-h)^2 + k$. The vertex of its graph is at $(h,k)$. The graph of the quadratic function opens upward when the coefficient, $a$, is positive and opens downward when $a$ is negative.

Not surprisingly, vertex form is especially helpful for finding the vertex of a graph of a quadratic function. For example, we can tell that the function, $p$, given by $p(x) = (x-3)^2 + 1$ has a vertex at $(3,1)$.

We also noticed that, when the squared expression $(x-3)^2$ has a positive coefficient, the graph opens upward. This means that the vertex, $(3,1)$, represents the minimum function value, $p(x)$.

Parabola in x y plane, origin O. X axis 0 to 6, by 1’s. Y axis 0 to 10, by 2s. Parabola opens upward with vertex at 3 comma 1. Left side of parabola crosses the y axis at 10.

But why does function $p$ take on its minimum value when $x$ is 3?

Here is one way to explain it: When $x=3$, the squared term $(x-3)^2$ equals 0, because $(3-3)^2 =0^2=0$. When $x$ is any other value besides 3, the squared term $(x-3)^2$ is a positive number greater than 0. (Squaring any number results in a positive number.) This means that the output when $x \neq 3$ will always be greater than the output when $x=3$, so function $p$ has a minimum value at $x=3$.

This table shows some values of the function for some values of $x$. Notice that the output is the least when $x=3$, and it increases both as $x$ increases and as it decreases.

The squared term sometimes has a negative coefficient, for instance in $h(x)= \text-2(x+4)^2$. The $x$ value that makes $(x+4)^2$ equal 0 is -4, because $(\text-4+4)^2=0^2=0$. Any other $x$ value makes $(x+4)^2$ greater than 0. But when $(x+4)^2$ is multiplied by a negative number like -2, the resulting expression, $\text-2(x+4)^2$, ends up being negative. This means that the output when $x \neq  \text-4$ will always be less than the output when $x=\text-4$, so function $h$ has its maximum value when $x =\text-4$.

Remember that we can find the $y$-intercept of the graph representing any function that we have seen. The $y$-coordinate of the $y$-intercept is the value of the function when $x = 0$. If $g$ is defined by $g(x)=(x+1)^2-5$, then the $y$-intercept is $(0,\text-4)$ because $g(0)=(0+1)^2 -5=\text-4$. Its vertex is at $(\text-1,\text-5)$. Another point on the graph with the same $y$-coordinate is located the same horizontal distance from the vertex but on the other side.

Parabola in x y plane, origin O. X axis negative 4 to 1, by 1’s. Y axis negative 8 to 2, by 2s. Parabola opens upward with vertex at negative 1 comma negative 5. Other points on the parabola given are negative 2 comma negative 4 and 0 comma negative 4.