2.1: Quadratic Functions
Learning objectives
 Learn how to solve a quadratic equation by the three methods: factoring, completing the square, and the quadratic formula.
 Understand the properties of quadratic functions and how to use them to make a graph.
Standard Form of Quadratic Function
We know from physics, if you throw a ball up in the air from, say, 10 meters above ground with an initial speed of 12 meters per second, then its height \begin{align*}h\end{align*} above the ground \begin{align*}t\end{align*} seconds later is given by the function \begin{align*}h(t)=4.9t^{2}+12t+10\end{align*}. This function is an example of a quadratic function.
Definition: Quadratic Function
A function \begin{align*}f\end{align*} defined by \begin{align*}f(x)=ax^{2}+bx+c\end{align*}, where \begin{align*}a, b,\end{align*} and \begin{align*}c\end{align*} are real numbers and \begin{align*}a \ne 0\end{align*}, is called a quadratic function.
The defining characteristic of a quadratic function are that it is a polynomial whose highest exponent is 2.
The form of the quadratic function above, \begin{align*}f(x)=ax^{2}+bx+c\end{align*}, is called the standard form of a quadratic function. There are several other ways to write quadratic functions such as in vertex form, \begin{align*}f(x)=a(xh)^{2}+k\end{align*}, and in factored form, \begin{align*}f(x)=a(xr_{1})(xr_{2})\end{align*}. You can move between forms of quadratic functions using algebra and you will see in this chapter that we can use each of these forms to graph a quadratic functions.
The \begin{align*}y\end{align*}intercept of a quadratic function in standard form is \begin{align*}(0, c)\end{align*} and it is found by substituting 0 for \begin{align*}x\end{align*} in \begin{align*}f(x)=ax^{2}+bx+c\end{align*}.
Graph Quadratic Functions Using Transformations
The shape of the graph of any quadratic function is called a parabola, and it is the same as the shape of \begin{align*}f(x)=x^{2}\end{align*} (see below). However, the basic graph may be moved, reflected, or stretched depending on the values of the coefficients \begin{align*}a, b,\end{align*} and \begin{align*}c\end{align*}. Examples of graphs of quadratic functions are shown in below. Notice that they all have the same shape, but some parabolas open upward and have a minimum point (called the vertex point) and some open downward and have a maximum vertex point. In calculus, those minimum and maximum points are usually called the extreme points of a function. As the name implies, at its extreme point(s), a function has a maximum or a minimum height.
Graph  Equation 

\begin{align*}f(x)=x^{2}\end{align*}  
\begin{align*}f(x)=x^{2}+2\end{align*}  
\begin{align*}f(x)=x^{2}+2x\end{align*}  
\begin{align*}f(x)=x^{2}6x+4\end{align*}  
\begin{align*}f(x)=x^{2}x+4\end{align*} 
One way to graph a quadratic function by hand is to translate the basic graph of \begin{align*}f(x)=x^{2}\end{align*} using transformations. You can apply rules about vertical shift, horizontal shift, and “stretching” of the basic parabolic shape. This is simplest to do if the quadratic function is written in vertex form by completing the square ((insert cross reference?)).
Summary of Vertex Form:
Given a quadratic function in the form \begin{align*}f(x)=a(xh)^{2}+k\end{align*}:
 The vertex is at \begin{align*}(h, k)\end{align*}
 The parabola opens up if \begin{align*}a>0\end{align*}
 The parabola opens down if \begin{align*}a<0\end{align*}
 The parabola is narrower than \begin{align*}y=x^{2}\end{align*} if \begin{align*}a>1\end{align*}
 The parabola opens wider than \begin{align*}y=x^{2}\end{align*} if \begin{align*}a<1\end{align*}
Example 1
Graph \begin{align*}g(x)=x^{2}+6x+7\end{align*} using transformations
Solution:
First we need to complete the square to write this function in vertex form. Add and subtract \begin{align*}\left ( \frac{b}{2} \right )^{2}\end{align*} to the right hand side of the equation:
\begin{align*}g(x) & = x^2 + 6x +7\\ & = x^2 + 6x + 9 + 7  9\end{align*}
Now factor the right hand side:
\begin{align*}g(x) & = (x+3)^2 2\end{align*}
Thus, \begin{align*}a=1\end{align*} and the vertex of this parabola is (3, 2). We know that the parabola opens up with the same width as \begin{align*}y=x^{2}\end{align*} and it has a minimum value at the vertex. the graph of the parabola is below.
Graph Quadratic Functions Using Vertex, Axis, Intercepts
Vertex of a parabola
Recall that the graph of a quadratic function is a parabola. In the the standard form of a quadratic function the \begin{align*}x\end{align*}coordinate of the vertex of the parabola is given by the equation
\begin{align*}x=\frac{b}{2a}\end{align*}
The \begin{align*}y\end{align*}coordinate of the vertex is found with
\begin{align*}y=f \left ( \frac{b}{2a} \right )\end{align*}
Axis of Symmetry of a Parabola
A parabola has reflective symmetry about a vertical line through the vertex. The vertical line \begin{align*}x=\frac{b}{2a}\end{align*} is also the parabola's axis of symmetry.
Example 2
Find the vertex and graph the quadratic function \begin{align*}g(x)=x^{2}8x+12\end{align*}
Solution: The \begin{align*}x\end{align*}coordinate of the vertex is \begin{align*}x=\frac{8}{2}=4\end{align*}.
The \begin{align*}y\end{align*}coordinate of the vertex is \begin{align*}g(4)=(4)^{2}8(4)+12=1632+12=4\end{align*}. Thus the vertex is at (4, 4).
To graph the parabola, we will make a table of points staring with the \begin{align*}x\end{align*}coordinate of 4:
\begin{align*}x\end{align*}  \begin{align*}y=g(x)\end{align*} 

4  4 
5  3 
6  0 
7  5 
Now we can use the symmetry of \begin{align*}g(x)\end{align*} to fill in this table for \begin{align*}g(3)\end{align*}. Note that \begin{align*}g(3)=g(5)=3\end{align*}. Likewise, \begin{align*}g(2)=g(6)=0\end{align*}. The final graph is below.
Intercepts of Quadratic Functions
Just like linear functions, the \begin{align*}y\end{align*} and \begin{align*}x\end{align*}intercepts of a quadratic function can be calculated by setting \begin{align*}x=0\end{align*} for the \begin{align*}y\end{align*}intercept and setting \begin{align*}y=0\end{align*} for the \begin{align*}x\end{align*}intercept. For example, to locate the \begin{align*}y\end{align*}intercept, substitute \begin{align*}x=0\end{align*} in \begin{align*}f(x)=ax^{2}+bx+c\end{align*} to obtain \begin{align*}y=f(0)=c\end{align*}. The \begin{align*}x\end{align*}intercepts are located by setting \begin{align*}y=0\end{align*} and then solving the quadratic equation \begin{align*}ax^{2}+bx+c=0\end{align*}, which can be solved by several methods that you have already studied from your previous mathematics courses: the factoring, completing the square, or using the quadratic formula. As a reminder, the quadratic formula is
Quadratic Formula
Given a quadratic function \begin{align*}f(x)=ax^{2}+bx+c\end{align*}, the \begin{align*}x\end{align*}intercepts of the function are:
\begin{align*}x=\frac{b\pm\sqrt{b^{2}4ac}}{2a}\end{align*}
All three methods–factoring, completing the square, or using the quadratic formula–give the \begin{align*}x\end{align*}intercepts or the roots or the solution set or the zeros of the quadratic equation (you can tell this is an important concept since there are so many words for it!). The following example gives you a quick review of how to use those three methods.
Graphing Quadratic Functions
 The quadratic function is defined as \begin{align*}y=f(x)=ax^{2}+bx+c\end{align*} and has an extreme point (the vertex) at the \begin{align*}(x, y)\end{align*} coordinates
\begin{align*}\left( \frac{b}{2a}, f\left ( \frac{b}{2a}\right ) \right )\end{align*}
If \begin{align*}a>0\end{align*} (positive), then the extreme point is a minimum and the parabola opens upwards.
If \begin{align*}a<0\end{align*} (negative), then the extreme point is a maximum and the parabola opens downwards.
 Find the \begin{align*}x\end{align*} and \begin{align*}y\end{align*}intercepts by setting \begin{align*}f(x)=0\end{align*}, and \begin{align*}f(0)=y\end{align*}, respectively.
Example 4
Sketch the graph of the function \begin{align*}y=f(x)=x^{2}+2x3\end{align*}.
Solution
Let's first find the intercepts. For the \begin{align*}y\end{align*}intercept, if \begin{align*}x=0\end{align*}, then \begin{align*}f(0)=3\end{align*}, or \begin{align*}y=3\end{align*}, so the \begin{align*}y\end{align*}intercept point is (0, 3).
Now, for the \begin{align*}x\end{align*}intercepts, if \begin{align*}y=f(x)=0\end{align*}, then \begin{align*}x^{2}+2x3=0\end{align*}, or
\begin{align*}x^{2}+2x3=(x+3)(x1)=0\end{align*}
so that \begin{align*}x=3\end{align*} and \begin{align*}x=1\end{align*} are the \begin{align*}x\end{align*}intercepts, that is (3, 0) and (1, 0).
The vertex (extreme point) is at
\begin{align*}x=\frac{b}{2a}=\frac{2}{2(1)}=1\end{align*}
Since
\begin{align*}f(1) & = (1)^2 + 2(1)3\\ & = 4\end{align*}
The vertex is (1, 4).
Since the coefficient of \begin{align*}x^{2}\end{align*} is positive, \begin{align*}a>0\end{align*}, the extreme point is a minimum and the parabola opens up. From this information, we can make a rough sketch of the parabola containing the points determined above. Notice that the range of the function is \begin{align*}y\ge4\end{align*}.
Example 5
Sketch the graph of the quadratic function \begin{align*}f(x)=x^{2}+4x\end{align*}.
Solution
To find the \begin{align*}y\end{align*}intercept, set \begin{align*}x=0\end{align*}, and \begin{align*}f(0)=(0)^{2}+4(0)=0\end{align*}. Thus the parabola intercepts the \begin{align*}y\end{align*}axis at the origin. The \begin{align*}x\end{align*}intercept is obtained by setting \begin{align*}y=0\end{align*}, thus \begin{align*}x^{2}+4x=0\end{align*}.
Factoring,
\begin{align*}x^{2}+4x = x(x4)=0\end{align*}
so that \begin{align*}x=0\end{align*} and \begin{align*}x=4\end{align*} are the \begin{align*}x\end{align*}intercepts.
We have \begin{align*}a=1\end{align*} and \begin{align*}b=4\end{align*}, so that the extreme point occurs when
\begin{align*}x=\frac{b}{2a}=\frac{4}{2(1)}=2\end{align*}
Since \begin{align*}f(2)=(2)^{2}+4(2)=4+8=4\end{align*}, then (2, 4) is the extreme point. It is a maximum point since \begin{align*}a=1<0\end{align*} and the parabola opens down. Finally, the graph can be obtained by sketching a parabola through the points determined above. From graph, the range of the function is \begin{align*}y \le 4\end{align*}.
Example 6
A projectile is shot from the ground level. If \begin{align*}h(t)=121t4.9t^{2}\end{align*}, where \begin{align*}t\end{align*} is the time in seconds and \begin{align*}h(t)\end{align*} is the height of the projectile (in meters) above the ground at time \begin{align*}t\end{align*}. Find (a) \begin{align*}h\end{align*} when \begin{align*}t=1\end{align*} second, (b) the maximum height reached by the projectile, and (c) the graph of the trajectory of the projectile.
Solution
(a) At \begin{align*}t=1\end{align*} second, we have \begin{align*}h(1)=121(1)4.9(1)^{2}=1214.9=116.1 \ meters\end{align*}.
(b) To find the maximum height reached by the projectile, we use the formula \begin{align*}x=\frac{b}{2a}\end{align*} to find the time at which the maximum height occurs (but here we use \begin{align*}t\end{align*} instead of \begin{align*}x\end{align*} for the independent variable):
\begin{align*}t=\frac{(121)}{2(4.9)}=12.3 \ seconds\end{align*}
In other words, the maximum height is reached at time 12.3 seconds. That tells us when the projectile reached its maximum height. To find the actual maximum height, we simply substitute this value into the height equation
\begin{align*}h(12.3)=121(12.3)4.9(12.3)^{2}=747 \ meters\end{align*}
(c) In order to graph it accurately, we need to find the \begin{align*}t\end{align*}intercepts. They can be found by setting \begin{align*}h=0\end{align*}:
\begin{align*}121t4.9t^{2} & = 0\\ (1214.9t)t & = 0\end{align*}
So the intercepts are at \begin{align*}t=0\end{align*} seconds and \begin{align*}t=24.7\end{align*} seconds. The \begin{align*}h\end{align*}intercept occurs when \begin{align*}t=0\end{align*}, so \begin{align*}h(0)=121(0)4.9(0)^{2}=0\end{align*}. From this information, we can construct the graph, as shown in the Figure 5. Notice that the graph does not extend beyond the interval on the \begin{align*}t\end{align*}axis. (why?)
Summarize Analysis of Quadratic Functions
Example 7
Find the roots (or \begin{align*}x\end{align*}intercepts) of the quadratic equation \begin{align*}x^{2}5x+6=0\end{align*} by using the three methods mentioned above.
Solution
1. The Factor Method
The factor method is based on writing the quadratic equation in factored form. That is, as a product of two linear expressions. So our equation maybe solved by the following way:
\begin{align*}x^{2}5x+6 & = 0\\ (x3)(x2) & = 0\end{align*}
Recall, that
\begin{align*}a \cdot b=0 \ \text{if and only if} \ a=0 \ \text{and} \ b=0\end{align*}
This tells us that
\begin{align*}x3=0\end{align*}
or
\begin{align*}x2=0\end{align*}
which give the roots (or zeros)
\begin{align*}x=3\end{align*}
and
\begin{align*}x=2\end{align*}
In other words, the solution set is {2, 3}.
2. Completing the Square Method
To solve the above equation by completing the square, first move the “\begin{align*}c\end{align*}” term to the other side of the equation,
\begin{align*}x^{2}5x=6\end{align*}
Next, make the lefthand side a “perfect square” by adding the appropriate number. To do so, take onehalf of the coefficient of \begin{align*}x\end{align*} (the \begin{align*}b\end{align*} coefficient) and square it and then add the result to both sides of the equation:
\begin{align*}b & = 5 \\ \frac{b}{2} & = \frac{5}{2}\\ \left ( \frac{b}{2} \right ) ^2 & = \frac{25}{4}\end{align*}
Adding to both sides of the equation,
\begin{align*}x^{2}5x+\frac{25}{4} & = 6 + \frac{25}{4}\\ x^2 5x + \frac{25}{4} & = \frac{1}{4}\\ \left ( x\frac{5}{2} \right )^2 & = \frac{1}{4}\end{align*}
this last equation can be easily solved by taking the square root of both sides,
\begin{align*}\left(x\frac{5}{2}\right)=\sqrt{\frac{1}{4}}=\pm\frac{1}{2}\end{align*}
Hence
\begin{align*}x=\pm\frac{1}{2}+\frac{5}{2}\end{align*}
and the solutions are
\begin{align*}x=3\end{align*}
and
\begin{align*}x=2\end{align*}
which are identical to answers of the factor method.
3. The Quadratic Formula
The quadratic formula is the most useful method in applications for solving quadratic functions because it works for all quadratic equations. Therefore, it is highly encouraged that you memorize the formula.
Recall, if \begin{align*}ax^{2}+bx+c=0\end{align*} where \begin{align*}a, b,\end{align*} and \begin{align*}c\end{align*} are real numbers and \begin{align*}a\ne0\end{align*}, then the roots of the equation can be determined by the quadratic formula
\begin{align*}x=\frac{b\pm\sqrt{b^{2}4ac}}{2a}.\end{align*}
Here the coefficients are \begin{align*}a=1, b=5,\end{align*} and \begin{align*}c=6\end{align*}.
Substituting into the quadratic formula, we get
\begin{align*}x & = \frac{(5) \pm \sqrt{(5)^2  4(1)(6)}}{2(1)}\\ & = \frac{5 \pm \sqrt{2524}}{2}\\ & = \frac{5 \pm 1}{2}\\ & = 3 \ \text{or} \ 2\end{align*}
which is, again, identical to our two solutions above.
Applications, Technological Tools
You can graph quadratic functions using your computer's graphing program or with a TI83/84 Calculator. Below we give basic directions for graphing a quadratic function and finding the vertex and roots (\begin{align*}x\end{align*}intercepts) using the functions on a TI83 or TI84 Graphing calculator.
Graphing
 As with graphing any function on the calculator, you enter the function's equation by using the \begin{align*}Y=\end{align*} button.
 Set an appropriate window using the WINDOW menu. You can use the vertex and \begin{align*}y\end{align*}intercept to help you find the best values for XMAX, XMIN, YMAX, YMIN, and the \begin{align*}X\end{align*} and \begin{align*}Y\end{align*}scales for viewing the graph of the parabola. Alternatively, you can set a standard window and use zooming to view the graph.
 Press GRAPH.
 Adjust and refine the view using ZOOM or by changing the WINDOW settings.
((Note: Insert 3 screen shots: 1) the \begin{align*}Y=\end{align*} menu with a quadratic function in \begin{align*}Y_{1}\end{align*}, 2) The WINDOW menu, and 3) A view of the graph.))
Finding the Vertex
You can use functions built into the calculator to find the vertex of any parabola you graph.
 Follow the directions above to graph a quadratic function.
 From the Graph screen, press 2ND TRACE (This is the CALC Menu).
 Scroll down and choose MINIMUM or MAXIMUM depending on whether the vertex is a minimum or maximum.
 The calculator takes you to the graph and prompts LEFT BOUND? Use the arrow keys (< or >) to place the cursor to the left of the vertex (or enter an \begin{align*}x\end{align*}coordinate by typing one in), and press ENTER
 The calculator then prompts RIGHT BOUND? Use the arrows to place the cursor to the right of the vertex and press ENTER.
 Finally, the calculator will prompt you for a guess with GUESS? You can either enter an \begin{align*}x\end{align*}value close to the vertex, or press ENTER again if your bounds were relatively good.
 The calculator will display the \begin{align*}x\end{align*} and \begin{align*}y\end{align*}coordinates of the vertex.
((Note: Add 2 screen shots: 1) the CALC menu in step #2, 2) A graph with LEFTBOUND? prompt (step #4), 3) A graph showing the vertex as MAX/MIN (i.e. after doing step 7)))
Finding the Roots
 From the Graph screen, press 2ND TRACE (This is the CALC Menu).
 Choose ZERO.
 The calculator takes you to the graph and prompts LEFT BOUND? Use the arrow keys (< or >) to place the cursor to the left of the zero (or enter an \begin{align*}x\end{align*}coordinate by typing one in), and press ENTER.
 The calculator then prompts RIGHT BOUND? Use the arrows to place the cursor to the right of the zero and press ENTER.
 Finally, the calculator will prompt you for a guess with GUESS? You can either enter an \begin{align*}x\end{align*}value close to the zero, or press ENTER again if your bounds were relatively good.
 The calculator will display the \begin{align*}x\end{align*}coordinate of the zero (root).
Note: Repeat steps 16 to find the other root of the quadratic.
((Note: Add 3 screen shots showing the calculator after Step 2, Step 4, and Step 6.))
Exercises
 Solve each equation by factoring.
 \begin{align*}x^{2}=64\end{align*}
 \begin{align*}4x^{2}+7x=2\end{align*}
 \begin{align*}4x^{2}17x=4\end{align*}
 \begin{align*}x^{2}+1=\frac{13}{6}x\end{align*}
 Solve each equation by completing the square
 \begin{align*}x^{2}4x+1=0\end{align*}
 \begin{align*}\frac{2}{3}x^{2}x+\frac{1}{3}=0\end{align*}
 \begin{align*}x^{2}+2.8=4.7x\end{align*}
 \begin{align*}2x^{2}3x\frac{23}{8}=0\end{align*}
 Solve each equation by using the quadratic formula
 \begin{align*}0.4x^{2}+x0.3=0\end{align*}
 \begin{align*}25x^{2}+80x+61=0\end{align*}
 \begin{align*}(z+6)^{2}+2z=0\end{align*}
 \begin{align*}\left(\frac{5}{7}w14\right)^{2}=8w\end{align*}
 Sketch the graph of the quadratic function \begin{align*}f(x)=x^{2}4x4\end{align*} by locating the \begin{align*}x\end{align*} and \begin{align*}y\end{align*}intercepts and the vertex point. Use the graph to determine the range of the parabola.
 Sketch the graph of the quadratic function \begin{align*}f(x)=2x^{2}+4x3\end{align*} by locating the \begin{align*}x\end{align*} and \begin{align*}y\end{align*}intercepts and the vertex point. Use the graph to determine the range of the parabola.
 Sketch the graph of the quadratic function \begin{align*}f(x)=x^{2}+x+5\end{align*} by locating the \begin{align*}x\end{align*} and \begin{align*}y\end{align*}intercepts and the vertex point. Use the graph to determine the range of the parabola.
 A projectile is fired from a cliff. If the height, \begin{align*}h\end{align*} of the projectile after \begin{align*}t\end{align*} seconds is given by \begin{align*}h(t)=16t^{2}+96t+256\end{align*}, find
 \begin{align*}h\end{align*} when \begin{align*}t=0\end{align*}.
 The maximum height reached by the projectile.
 Graph \begin{align*}h(t)\end{align*} and show its range and domain.
 A surfacetosurface missile is fired and follows a parabolic path. Its path as a function of time is described by the function \begin{align*}p(t)=t^{2}+t+1\end{align*}. At what time the missile reaches the highest point and when does it hit the ground?
 Use a graphing calculator to find the vertex and roots of \begin{align*}q(x)=0.035x^{2}7.25x12.3\end{align*}. Round your answer to three decimal places.
Answers

 \begin{align*}\pm8\end{align*}
 \begin{align*}x=\frac{1}{4}, 2\end{align*}
 \begin{align*}x=\frac{1}{4}, 4\end{align*}
 \begin{align*}x=\frac{2}{3}, \frac{3}{2}\end{align*}
 \begin{align*}2\pm\sqrt{3}\end{align*}
 \begin{align*}x=\frac{1}{2}, 1\end{align*}
 \begin{align*}x=\frac{7}{10}, 4\end{align*}
 \begin{align*}\frac{3}{4}\pm\sqrt{2}\end{align*}
 \begin{align*}x=0.275, 2.78\end{align*}
 \begin{align*}\frac{8}{5}\pm\frac{\sqrt{3}}{5}\end{align*}
 \begin{align*}7\pm\sqrt{13}\end{align*}
 \begin{align*}\frac{686\pm196\sqrt{6}}{25}\end{align*}
 The vertex and the \begin{align*}x\end{align*}intercept is (2, 0). The \begin{align*}y\end{align*}intercept is (0, 4).
 Vertex: (1, 5). \begin{align*}x\end{align*}intercepts: \begin{align*}\left(\frac{2+\sqrt{10}}{2}, 0\right)\end{align*} and \begin{align*}\left(\frac{2\sqrt{10}}{2}, 0\right)\end{align*}. \begin{align*}y\end{align*}intercept: (0, 3).
 Vertex: \begin{align*}\left(\frac{1}{2}, \frac{19}{4}\right)\end{align*}. No \begin{align*}x\end{align*}intercepts. \begin{align*}y\end{align*}intercept: (0, 5)
 256 ft
 400 ft
 At \begin{align*}\frac{1}{2}\end{align*} second; after 1 second
 Vertex: (103.571, 363.146); roots at (205.432, 0) and (1.711, 0).
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