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Section 1.5 Quadratic Functions

After linear functions, quadratic functions are arguably the next simplest functions in mathematics. A quadratic function is one that may be written in the form

\begin{equation*} q(x) = ax^2 + bx + c\text{,} \end{equation*}

where \(a\text{,}\) \(b\text{,}\) and \(c\) are real numbers with \(a \ne 0\text{.}\) One of the reasons that quadratic functions are especially important is that they model the height of an object falling under the force of gravity.

Preview Activity 1.5.1.

A water balloon is tossed vertically from a fifth story window. Its height, \(h\text{,}\) in meters, at time \(t\text{,}\) in seconds, is modeled by the function

\begin{equation*} h = q(t) = -5t^2 + 20t + 25\text{.} \end{equation*}
  1. Execute appropriate computations to complete both of the following tables.

    Table 1.5.1. Function values for \(h\) at select inputs.
    \(t\) \(h = q(t)\)
    \(0\) \(q(0) = 25\)
    \(1\)
    \(2\)
    \(3\)
    \(4\)
    \(5\)
    Table 1.5.2. Average rates of change for \(h\) on select intervals.
    \([a,b]\) \(AV_{[a,b]}\)
    \([0,1]\) \(AV_{[0,1]} = 15\) m/s
    \([1,2]\)
    \([2,3]\)
    \([3,4]\)
    \([4,5]\)
  2. What pattern(s) do you observe in Tables 1.5.1 and 1.5.2?

  3. Explain why \(h = q(t)\) is not a linear function. Use Definition 1.4.3 in your response.

  4. What is the average velocity of the water balloon in the final second before it lands? How does this value compare to the average velocity on the time interval \([4.9, 5]\text{?}\)

Subsection 1.5.1 Properties of Quadratic Functions

Quadratic functions are likely familiar to you from experience in previous courses. Throughout, we let \(y = q(x) = ax^2 + bx + c\) where \(a\text{,}\) \(b\text{,}\) and \(c\) are real numbers with \(a \ne 0\text{.}\) From the outset, it is important to note that when we write \(q(x) = ax^2 + bx + c\) we are thinking of an infinite family of functions where each member depends on the three paramaters \(a\text{,}\) \(b\text{,}\) and \(c\text{.}\)

Activity 1.5.2.

Open a browser and point it to Desmos. In Desmos, enter q(x) = ax^2 + bx + c; you will be prompted to add sliders for \(a\text{,}\) \(b\text{,}\) and \(c\text{.}\) Do so. Then begin exploring with the sliders and respond to the following questions.

  1. Describe how changing the value of \(a\) impacts the graph of \(q\text{.}\)

  2. Describe how changing the value of \(b\) impacts the graph of \(q\text{.}\)

  3. Describe how changing the value of \(c\) impacts the graph of \(q\text{.}\)

  4. Which parameter seems to have the simplest effect? Which parameter seems to have the most complicated effect? Why?

  5. Is it possible to find a formula for a quadratic function that passes through the points \((0,8)\text{,}\) \((1,12)\text{,}\) \((2,12)\text{?}\) If yes, do so; if not, explain why not.

Because quadratic functions are familiar to us, we will quickly restate some of their important known properties.

Solutions to \(q(x) = 0\).

Let \(a\text{,}\) \(b\text{,}\) and \(c\) be real numbers with \(a \ne 0\text{.}\) The equation \(ax^2 + bx + c = 0\) can have \(0\text{,}\) \(1\text{,}\) or \(2\) real solutions. These real solutions are given by the quadratic formula,

\begin{equation*} x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\text{,} \end{equation*}

provided that \(b^2 - 4ac \ge 0\text{.}\)

As we can see in Figure 1.5.3, by shifting the graph of a quadratic function vertically, we can make its graph cross the \(x\)-axis \(0\) times (as in the graph of \(p\)), exactly \(1\) time (\(q\)), or twice (\(r\)). These points are the \(x\)-intercepts of the graph.

Figure 1.5.3. Three examples of quadratic functions that open up.
Figure 1.5.4. One example of a quadratic function that opens down.

While the quadratic formula will always provide any real solutions to \(q(x) = 0\text{,}\) in practice it is often easier to attempt to factor before using the formula. For instance, given \(q(x) = x^2 - 5x + 6\text{,}\) we can find its \(x\)-intercepts quickly by factoring. Since

\begin{equation*} x^2 - 5x + 6 = (x-2)(x-3)\text{,} \end{equation*}

it follows that \((2,0)\) and \((3,0)\) are the \(x\)-intercepts of \(q\text{.}\) Note more generally that if we know the \(x\)-intercepts of a quadratic function are \((r,0)\) and \((s,0)\text{,}\) it follows that we can write the quadratic function in the form \(q(x) = a(x-r)(x-s)\text{.}\)

Every quadratic function has a \(y\)-intercept; for a function of form \(y = q(x) = ax^2 + bx + c\text{,}\) the \(y\)-intercept is the point \((0,c)\text{,}\) as demonstrated in Figure 1.5.4.

In addition, every quadratic function has a symmetric graph that either always curves upward or always curves downward. The graph opens upward if and only if \(a \gt 0\) and opens downward if and only if \(a \lt 0\text{.}\) We often call the graph of a quadratic function a parabola. Every parabola is symmetric about a vertical line that runs through its lowest or highest point.

The vertex of a parabola.

The quadratic function \(y = q(x) = ax^2 + bx + c\) has its vertex at the point \(\left( -\frac{b}{2a}, q\left( -\frac{b}{2a} \right) \right)\text{.}\) When \(a \gt 0\text{,}\) the vertex is the lowest point on the graph of \(q\text{,}\) while if \(a \lt 0\text{,}\) the vertex is the highest point. Moreover, the graph of \(q\) is symmetric about the vertical line \(x = -\frac{b}{2a}\text{.}\)

Figure 1.5.5. The vertex of a quadratic function that opens up.
Figure 1.5.6. The vertex of a quadratic function that opens down.

Note particularly that due to symmetry, the vertex of a quadratic function lies halfway between its \(x\)-intercepts (provided the function has \(x\)-intercepts). In both Figures 1.5.5 and 1.5.6, we see how the parabola is symmetric about the vertical line that passes through the vertex. One way to understand this symmetry can be seen by writing a given quadratic function in a different algebraic form.

Example 1.5.7.

Consider the quadratic function in standard form given by \(y = q(x) = 0.25x^2 - x + 3.5\text{.}\) Determine constants \(a\text{,}\) \(h\text{,}\) and \(k\) so that \(q(x) = a(x-h)^2 + k\text{,}\) and hence determine the vertex of \(q\text{.}\) How does this alternate form of \(q\) explain the symmetry in its graph?

Solution

We first observe that we can write \(q(x) = 0.25x^2 - x + 3.5\) in a form closer to \(q(x) = a(x-h)^2 + k\) by factoring \(0.25\) from the first two terms to get

\begin{equation*} q(x) = 0.25(x^2 - 4x) + 3.5\text{.} \end{equation*}

Next, we want to add a constant inside the parentheses to form a perfect square. Noting that \((x-2)^2 = x^2 - 4x + 4\text{,}\) we need to add \(4\text{.}\) Since we are adding \(4\) inside the parentheses, the \(4\) is being multiplied by \(0.25\text{,}\) which has the net effect of adding \(1\) to the function. To keep the function as given, we must also subtract \(1\text{,}\) and thus we have

\begin{equation*} q(x) = 0.25(x^2 - 4x + 4) + 3.5 - 1\text{.} \end{equation*}

It follows that

\begin{equation*} q(x) = 0.25(x-2)^2 + 2.5\text{.} \end{equation*}

Next, observe that the vertex of \(q\) is \((2,2.5)\text{.}\) This holds because \((x-2)^2\) is always greater than or equal to \(0\text{,}\) and thus its smallest possible value is \(0\) when \(x = 2\text{.}\) Moreover, when \(x = 2\text{,}\) \(q(2) = 2.5\text{.}\) 1 

We can also verify this point is the vertex using standard form. From \(q(x) = 0.25x^2 - x + 3.5\text{,}\) we see that \(a = 0.25\) and \(b = -1\text{,}\) so \(x = -\frac{b}{2a} = \frac{1}{0.5} = 2\text{.}\) In addition, \(q(2) = 2.5\text{.}\)

Finally, the form \(q(x) = 0.25(x-2)^2 + 2.5\) explains the symmetry of \(q\) about the line \(x = 2\text{.}\) Consider the two points that lie equidistant from \(x = 2\) on the \(x\)-axis, \(z\) units away: \(x = 2-z\) and \(x = 2 + z\text{.}\) Observe that for these values,

\begin{align*} q(2-z) &= 0.25(2-z-2)^2 + 2.5& q(2+z) &= 0.25(2+z-2)^2 + 2.5\\ &= 0.25(-z)^2 + 2.5 & &= 0.25(z)^2 + 2.5\\ &= 0.25z^2 + 2.5 & &= 0.25z^2 + 2.5 \end{align*}

Since \(q(2-z) = q(2+z)\) for any choice of \(z\text{,}\) this shows the parabola is symmetric about the vertical line through its vertex.

In Example 1.5.7, we saw some of the advantages of writing a quadratic function in the form \(q(x) = a(x-h)^2 + k\text{.}\) We call this the vertex form of a quadratic function.

Vertex form of a quadratic function.

A quadratic function with vertex \((h,k)\) may be written in the form \(y = a(x-h)^2 + k\text{.}\) The constant \(a\) may be determined from one other function value for an input \(x \ne h\text{.}\)

Activity 1.5.3.

Reason algebraically using appropriate properties of quadratic functions to answer the following questions. Use Desmos to check your results graphically.

  1. How many quadratic functions have \(x\)-intercepts at \((-5,0)\) and \((10,0)\) and a \(y\)-intercept at \((0,-1)\text{?}\) Can you determine an exact formula for such a function? If yes, do so. If not, explain why.

  2. Suppose that a quadratic function has vertex \((-3,-4)\) and opens upward. How many \(x\)-intercepts can you guarantee the function has? Why?

  3. In addition to the information in (b), suppose you know that \(q(-1) = -3\text{.}\) Can you determine an exact formula for \(q\text{?}\) If yes, do so. If not, explain why.

  4. Does the quadratic function \(p(x) = -3(x+1)^2 + 9\) have \(0\text{,}\) \(1\text{,}\) or \(2\) \(x\)-intercepts? Reason algebraically to determine the exact values of any such intercepts or explain why none exist.

  5. Does the quadratic function \(w(x) = -2x^2 + 10x - 20\) have \(0\text{,}\) \(1\text{,}\) or \(2\) \(x\)-intercepts? Reason algebraically to determine the exact values of any such intercepts or explain why none exist.

Subsection 1.5.2 Modeling falling objects

One of the reasons that quadratic functions are so important is because of a physical fact of the universe we inhabit: for an object only being influenced by gravity, acceleration due to gravity is constant. If we measure time in seconds and a rising or falling object's height in feet, the gravitational constant is \(g = -32\) feet per second per second.

One of the fantastic consequences of calculus — which, like the realization that acceleration due to gravity is constant, is largely due to Sir Isaac Newton in the late 1600s — is that the height of a falling object at time \(t\) is modeled by a quadratic function.

Height of an object falling under the force of gravity.

For an object tossed vertically from an initial height of \(s_0\) feet with a velocity of \(v_0\) feet per second, the object's height at time \(t\) (in seconds) is given by the formula

\begin{equation*} h(t) = -16t^2 + v_0t + s_0 \end{equation*}

If height is measured instead in meters and velocity in meters per second, the gravitational constant is \(g = 9.8\) and the function \(h\) has form \(h(t) = -4.9t^2 + v_0t + s_0\text{.}\) (When height is measured in feet, the gravitational constant is \(g = 32\text{.}\))

Activity 1.5.4.

A water balloon is tossed vertically from a window at an initial height of 37 feet and with an initial velocity of 41 feet per second.

  1. Determine a formula, \(s(t)\text{,}\) for the function that models the height of the water balloon at time \(t\text{.}\)

  2. Plot the function in Desmos in an appropriate window.

  3. Use the graph to estimate the time the water balloon lands.

  4. Use algebra to find the exact time the water balloon lands.

  5. Determine the exact time the water balloon reaches its highest point and its height at that time.

  6. Compute the average rate of change of \(s\) on the intervals \([1.5, 2]\text{,}\) \([2, 2.5]\text{,}\) \([2.5,3]\text{.}\) Include units on your answers and write one sentence to explain the meaning of the values you found. Sketch appropriate lines on the graph of \(s\) whose respective slopes are the values of these average rates of change.

Subsection 1.5.3 How quadratic functions change

So far, we've seen that quadratic functions have many interesting properties. In Preview Activity 1.5.1, we discovered an additional pattern that is particularly noteworthy.

Recall that we considered a water balloon tossed vertically from a fifth story window whose height, \(h\text{,}\) in meters, at time \(t\text{,}\) in seconds, is modeled 2  by the function

\begin{equation*} h = q(t) = -5t^2 + 20t + 25\text{.} \end{equation*}

We then completed Table 1.5.8 and Table 1.5.9 to investigate how both function values and averages rates of change varied as we changed the input to the function.

Here we are using \(a = -5\) rather than \(a = -4.9\) for simplicity.
Table 1.5.8. Function values for \(h\) at select inputs.
\(t\) \(h = q(t)\)
\(0\) \(q(0) = 25\)
\(1\) \(q(1) = 40\)
\(2\) \(q(2) = 45\)
\(3\) \(q(3) = 40\)
\(4\) \(q(4) = 25\)
\(5\) \(q(5) = 0\)
Table 1.5.9. Average rates of change for \(h\) on select intervals \([a,b]\text{.}\)
\([a,b]\) \(AV_{[a,b]}\)
\([0,1]\) \(AV_{[0,1]} = 15\) m/s
\([1,2]\) \(AV_{[1,2]} = 5\) m/s
\([2,3]\) \(AV_{[2,3]} = -5\) m/s
\([3,4]\) \(AV_{[3,4]} = -15\) m/s
\([4,5]\) \(AV_{[4,5]} = -25\) m/s

In Table 1.5.9, we see an interesting pattern in the average velocities of the ball. Indeed, if we remove the “\(AV\)” notation and focus on the starting value of each interval, viewing the resulting average rate of change, \(r\text{,}\) as a function of the starting value, we may consider the related table seen in Table 1.5.10, where it is apparent that \(r\) is a linear function of \(a\text{.}\)

Table 1.5.10. Data from Table 1.5.9, slightly recast.
\(a\) \(r(a)\)
\(0\) \(r(0) = 15\) m/s
\(1\) \(r(1) = 5\) m/s
\(2\) \(r(2) = -5\) m/s
\(3\) \(r(3) = -15\) m/s
\(4\) \(r(4) = -25\) m/s
Figure 1.5.11. Plot of \(h(t) = -5t^2 + 20t + 25\) along with line segments whose slopes correspond to average rates of change.

Indeed, viewing this data graphically as in Figure 1.5.11, we observe that the average rate of change of \(h\) is itself changing in a way that seems to be represented by a linear function. While it takes key ideas from calculus to formalize this observation, for now we will simply note that for a quadratic function there seems to be a related linear function that tells us something about how the quadratic function changes. Moreover, we can also say that on the downward-opening quadratic function \(h\) that its average rate of change appears to be decreasing as we move from left to right 3 .

Provided that we consider the average rate of change on intervals of the same length. Again, it takes ideas from calculus to make this observation completely precise.

A key closing observation here is that the fact the parabola “bends down” is apparently connected to the fact that its average rate of change decreases as we move left to right. By contrast, for a quadratic function that “bends up”, we can show that its average rate of change increases as we move left to right (see Exercise 1.5.5.7). Moreover, we also see that it's possible to view the average rate of change of a function on \(1\)-unit intervals as itself being a function: a process that relates an input (the starting value of the interval) to a corresponding output (the average rate of change of the original function on the resulting \(1\)-unit interval).

For any function that consistently bends either exclusively upward or exclusively downward on a given interval \((a,b)\text{,}\) we use the following formal language 4  to describe it.

Calculus is needed to make Definition 1.5.12 rigorous and precise.
Definition 1.5.12.

If a function \(f\) always bends upward on an interval \((a,b)\text{,}\) we say that \(f\) is concave up on \((a,b)\). Similarly, if \(f\) always bends downward on an interval \((a,b)\text{,}\) we say that \(f\) is concave down on \((a,b)\).

Thus, we now call a quadratic function \(q(x) = ax^2 + bx + c\) with \(a \gt 0\) “concave up”, while if \(a \lt 0\) we say \(q\) is “concave down”.

Subsection 1.5.4 Summary

  • Quadratic functions (of the form \(q(x) = ax^2 + bx + c\) with \(a \ne 0\)) are emphatically not linear: their average rate of change is not constant, but rather depends on the interval chosen. At the same time, quadratic functions appear to change in a very regimented way: if we compute the average rate of change on several consecutive \(1\)-unit intervals, it appears that the average rate of change itself changes at a constant rate. Quadratic functions either bend upward (\(a \gt 0\)) or bend downward (\(a \lt 0\)) and these shapes are connected to whether the average rate of change on consecutive \(1\)-unit intervals decreases or increases as we move left to right.

  • For an object with height \(h\) measured in feet at time \(t\) in seconds, if the object was launched vertically at an initial velocity of \(v_0\) feet per second and from an initial height of \(s_0\) feet, the object's height is given by

    \begin{equation*} h = q(t) = -16t^2 + v_0t + s_0\text{.} \end{equation*}

    That is, the object's height is completely determined by the initial height and initial velocity from which it was launched. The model is valid for the entire time until the object lands. If \(h\) is instead measured in meters and \(v_0\) in meters per second, \(-16\) is replaced with \(-4.9\text{.}\)

  • A quadratic function \(q\) can be written in one of three familiar forms: standard, vertex, or factored 5 . Table 1.5.13 shows how, depending on the algebraic form of the function, various properties may be (easily) read from the formula. In every case, the sign of \(a\) determines whether the function opens up or opens down.

    Table 1.5.13. A summary of the information that can be read from the various algebraic forms of a quadratic function
    standard vertex factored 6 
    form \(q(x) = ax^2 + bx + c\) \(q(x) = a(x-h)^2 + k\) \(q(x) = a(x-r)(x-s)\)
    \(y\)-int \((0,c)\) \((0,ah^2 + k)\) \((0,ars)\)
    \(x\)-int 7  \(\left(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a} , 0 \right)\) \(\left(h \pm \sqrt{-\frac{k}{a}} , 0 \right)\) \((r,0)\text{,}\) \((s,0)\)
    vertex \(\left(-\frac{b}{2a}, q\left( -\frac{b}{2a} \right) \right)\) \((h,k)\) \(\left( \frac{r+s}{2}, q\left( \frac{r+s}{2} \right) \right)\)
It's not always possible to write a quadratic function in factored form involving only real numbers; this can only be done if it has \(1\) or \(2\) \(x\)-intercepts.
Provided \(q\) has \(1\) or \(2\) \(x\)-intercepts. In the case of just one, we take \(r = s\text{.}\)
Provided \(b^2 - 4ac \ge 0\) for standard form; provided \(-\frac{k}{a} \ge 0\) for vertex form.

Exercises 1.5.5 Exercises

1.
2.
3.
4.
5.
6.

Two quadratic functions, \(f\) and \(g\text{,}\) are determined by their respective graphs in Figure 1.5.14.

Figure 1.5.14. Two quadratic functions, \(f\) and \(g\text{.}\)
  1. How does the information provided enable you to find a formula for \(f\text{?}\) Explain, and determine the formula.

  2. How does the information provided enable you to find a formula for \(g\text{?}\) Explain, and determine the formula.

  3. Consider an additional quadratic function \(h\) given by \(h(x) = 2x^2 - 8x + 6\text{.}\) Does the graph of \(h\) intersect the graph of \(f\text{?}\) If yes, determine the exact points of intersection, with justification. If not, explain why.

  4. Does the graph of \(h\) intersect the graph of \(g\text{?}\) If yes, determine the exact points of intersection, with justification. If not, explain why.

7.

Consider the quadratic function \(f\) given by \(f(x) = \frac{1}{2}(x-2)^2 + 1\text{.}\)

  1. Determine the exact location of the vertex of \(f\text{.}\)

  2. Does \(f\) have \(0\text{,}\) \(1\text{,}\) or \(2\) \(x\)-intercepts? Explain, and determine the location(s) of any \(x\)-intercept(s) that exist.

  3. Complete the following tables of function values and average rates of change of \(f\) at the stated inputs and intervals.

    Table 1.5.15. Function values for \(f\) at select inputs.
    \(x\) \(f(x)\)
    \(0\)
    \(1\)
    \(2\)
    \(3\)
    \(4\)
    \(5\)
    Table 1.5.16. Average rates of change for \(f\) on select intervals.
    \([a,b]\) \(AV_{[a,b]}\)
    \([0,1]\) \(\)
    \([1,2]\)
    \([2,3]\)
    \([3,4]\)
    \([4,5]\)
  4. What pattern(s) do you observe in Table 1.5.15 and 1.5.16?

8.

A water balloon is tossed vertically from a window on the fourth floor of a dormitory from an initial height of \(56.3\) feet. A person two floors above observes the balloon reach its highest point \(1.2\) seconds after being launched.

  1. What is the balloon's exact height at \(t = 2.4\text{?}\) Why?

  2. What is the exact maximum height the balloon reaches at \(t = 1.2\text{?}\)

  3. What exact time did the balloon land?

  4. At what initial velocity was the balloon launched?