Starting with “a” in the standard form

As the lead coefficient of the standard form of the quadratic function , a gives you two bits of information: the direction in which the graphed parabola opens, and whether the parabola is steep or flat. Here’s the breakdown of how the sign and size of the lead coefficient, a, affect the parabola’s appearance:

• If a is positive, the graph of the parabola opens upward (see Figures 7-1a and 7-1b).
• If a is negative, the graph of the parabola opens downward (see Figures 7-1c and 7-1d).
• If a has an absolute value greater than one, the graph of the parabola is steep (see Figures 7-1a and 7-1c). (See Chapter 2 for a refresher on absolute values.)
• If a has an absolute value less than one, the graph of the parabola flattens out (see Figures 7-1b and 7-1d).

If you remember the four rules that identify the lead coefficient of a parabola, you don’t even have to graph the equation to describe how the parabola looks. Here’s how you can describe some parabolas from their equations:

• : You say that this parabola is steep and opens upward because the lead coefficient is positive and greater than one.
• : You say that this parabola is flattened out and opens downward because the lead coefficient is negative, and the absolute value of the fraction is less than one.
• : You say that this parabola is flattened out and opens upward because the lead coefficient is positive, and the decimal value is less than one. In fact, the coefficient is so small that the flattened parabola almost looks like a horizontal line.

Following up with “b” and “c”

Much like the lead coefficient in the quadratic function (see the previous section), the values of b and c give you plenty of information. Mainly, the values tell you a lot if they’re not there. In the next section, you find out how to use these values to find intercepts (or zeros). For now, you concentrate on their presence or absence.

The lead coefficient, a, can never be equal to zero. If that happens, you no longer have a quadratic function, and this discussion is finished. As for the other two terms:

• If the second coefficient, b, is zero, the parabola straddles the y-axis. The parabola’s vertex — the highest or lowest point on the curve, depending on which way it faces — is on that axis, and the parabola is symmetric about the axis (see Figure 7-2a, a graph of a quadratic function where ). The second term in the standard form is the x term, so if the coefficient b is zero, the second term disappears. The standard equation becomes , which makes finding intercepts very easy (see the following section).
• If the last term, c, is zero, the graph of the parabola goes through the origin — in other words, one of its intercepts is the origin (see Figure 7-2b, a graph of a quadratic function where ). The standard equation becomes , which you can easily factor into . (See Chapters 1 and 3 for more on factoring.)

Finding the one and only y-intercept

The y-intercept of a quadratic function is . A parabola with the standard equation is a function, so by definition (as I cover in Chapter 6), only one y value can exist for every x value. When , as it does at the y-intercept, the equation becomes , or . The statements and combine to become the y-intercept, .

To find the y-intercepts of the following functions, you let :

• : When , (or ). The y-intercept is (0, 2).
• : When , (or ); don’t let the missing x term throw you. The y-intercept is .
• : When , . The equation provides no constant term; you could also say the missing constant term is zero. The y-intercept is (0, 0).

People can model many situations with quadratic functions, and the places where the input variables or output variables equal zero are important. For example, a candle-making company has figured out that its profit is based on the number of candles it produces and sells. The company uses the function — where x represents the number of candles — to determine P, the profit. As you can see from the equation, the graph of this parabola opens downward (because a is negative; see the section “Starting with ‘a’ in the standard form”). Figure 7-3 is a sketch of the graph of the profit, with the y-axis representing profit and the x-axis representing the number of candles.

Does it make sense to use a quadratic function to model profit? Why would the profit decrease after a certain point? Does that make business sense? It does if you consider that perhaps, when you make too many candles, the cost of overtime and the need for additional machinery play a part.

What about the y-intercept? What part does it play, and what does it mean in this candle-making case? You can say that represents not producing or selling any candles. According to the equation and graph, the y-intercept has a y-coordinate of . It makes sense to find a negative profit if the company has costs that it has to pay no matter what (even if it sells no candles): insurance, salaries, mortgage payments, and so on. With some interpretation, you can find a logical explanation for the y-intercept being negative in this case.

Finding the x-intercepts

You find the x-intercepts of quadratics when you solve for the zeros, or solutions, of the quadratic equation. The method you use to solve for the zeros is the same method you use to solve for the intercepts, because they’re really just the same thing. The names change (intercept, zero, solution), depending on the application, but you find the intercepts the same way.

Parabolas with an equation of the standard form open upward or downward and may or may not have x-intercepts. Look at Figure 7-4, for example. You see a parabola with two x-intercepts (Figure 7-4a), one with a single x-intercept (Figure 7-4b), and one with no x-intercept (Figure 7-4c). Notice, however, that they all have a y-intercept.

The coordinates of all x-intercepts have zeros in them. An x-intercept’s y value is zero, and you write it in the form (h, 0). How do you find the value of h? You let in the general equation and then solve for x. You have two options when solving the equation :

• Use the quadratic formula (refer to Chapter 3 for a refresher on the formula).
• Try to factor the expression and use the multiplication property of zero (you can find more on this in Chapter 1).

Regardless of the path you take, you have some guidelines at your disposal to help you determine the number of x-intercepts you should find.

When finding x-intercepts by solving ,

• You find two x-intercepts if
  • The expression factors into two different binomials.
  • The quadratic formula gives you a value greater than zero under the radical.
• You find one x-intercept (a double root) if
  • The expression factors into the square of a binomial.
  • The quadratic formula gives you a value of zero under the radical.
• You find no x-intercept if both
  • The expression doesn’t factor.
  • The quadratic formula gives you a value less than zero under the radical (indicating an imaginary root; Chapter 14 deals with imaginary and complex numbers).

To find the x-intercepts of , for example, you can set y equal to zero and solve the quadratic equation by factoring:

• 
• When , ; and when , .

The two intercepts are and ; you can see that the equation factors into two different factors. In cases where you can’t figure out how to factor the quadratic, you can get the same answer by using the quadratic formula. I show you how to do this, using the quadratic that I just factored. Notice in the following calculation that the value under the radical is a number greater than zero, meaning that you have two answers:

When performing the addition in the numerator, you have . And when performing the subtraction, . The answers are, of course, the same.

Here’s another example, with a different result. To find the x-intercepts of , you can set y equal to zero and solve the quadratic equation by factoring:

• 
• So when , .

The only intercept is (4, 0). The equation factors into the square of a binomial — a double root. You can get the same answer by using the quadratic formula — notice that the value under the radical is equal to zero:

This last example shows how you determine that an equation has no x-intercept. To find the x-intercepts of , you can set y equal to zero and try to factor the quadratic equation, but you can’t do it. The equation has no factors that give you this quadratic.

When you try the quadratic formula, you see that the value under the radical is less than zero; a negative number under the radical is an imaginary number:

Alas, you find no x-intercept for this parabola.

Selling candles

A candle-making company has figured out that its profit is based on the number of candles it produces and sells. The function applies to the company’s situation, where x represents the number of candles, and P represents the profit. You may recognize this function from the section “Investigating Intercepts in Quadratics” earlier in the chapter. You can use the function to find out how many candles the company has to produce to garner the greatest possible profit.

You find the two x-intercepts by letting y = 0 and solving for x by factoring:

• 
• When , you have ; and when , . These two numbers give you the two x-intercepts: (20, 0) and (140, 0).

The intercept (20, 0) represents where the function (the profit) changes from negative values to positive values. You know this because the graph of the profit function is a parabola that opens downward (because a is negative), so the beginning and ending of the curve appear below the x-axis. The intercept (140, 0) represents where the profit changes from positive values to negative values. So, the maximum value, the vertex, lies somewhere between and above the two intercepts (see the section “Going to the Extreme: Finding the Vertex”). The x-coordinate of the vertex lies between 20 and 140. Refer to Figure 7-3 if you want to see the graph again.

You now use the formula for the x-coordinate of the vertex to find . The number 80 lies between 20 and 140; in fact, it rests halfway between them. The nice, even number is due to the symmetry of the graph of the parabola and the symmetric nature of these functions. Now you can find the P value (the y-coordinate of the vertex): .

Your findings say that if the company produces and sells 80 candles, the maximum profit will be $180. That seems like an awful lot of work for $180, but maybe the company runs a small business. Work such as this shows you how important it is to have models for profit, revenue, and cost in business so you can make projections and adjust your plans.

Shooting basketballs

A local youth group recently raised money for charity by having a Throw-A-Thon. Participants prompted sponsors to donate money based on a promise to shoot baskets over a 12-hour period. This was a very successful project, both for charity and for algebra, because you can find some interesting bits of information about shooting the basketballs and the number of misses that occurred.

Participants shot baskets for 12 hours, attempting about 200 baskets each hour. The quadratic equation models the number of baskets they missed each hour, where t is the time in hours (numbered from 0 through 12) and M is the number of misses.

This quadratic function opens upward (because a is positive), so the function has a minimum value. Figure 7-8 shows a graph of the function.

From the graph, you see that the initial value, the y-intercept, is 100. At the beginning, participants were missing about 100 baskets per hour. The good news is that they got better with practice. , which means that at hour two into the project, the participants were missing only 60 baskets per hour. The number of misses goes down and then goes back up again. How do you interpret this? Even though the participants got better with practice, they let the fatigue factor take over.

What’s the fewest number of misses per hour? When did the participants shoot their best? To answer these questions, find the vertex of the parabola by using the formula for the x-coordinate (you can find this in the “Going to the Extreme: Finding the Vertex” section earlier in the chapter):

The best shooting happened about 4.5 hours into the project. How many misses occurred then? The number you get represents what’s happening the entire hour — although that’s fudging a bit:

The fraction is rounded to two decimal places. The best shooting is about 42 misses that hour.

Launching a water balloon

One of the favorite springtime activities of the engineering students at a certain university is to launch water balloons from the top of the engineering building so they hit the statue of the school’s founder, which stands 25 feet from the building. The launcher sends the balloons up in an arc to clear a tree that sits next to the building. To hit the statue, the initial velocity and angle of the balloon have to be just right. Figure 7-9 shows a successful launch.

This year’s launch was successful. The students found that by launching the water balloons at 48 feet per second with a precise angle, they could hit the statue. Here’s the equation they worked out to represent the path of the balloons: . The t represents the number of seconds, and H is the height of the balloon in feet. From this quadratic function, you can answer the following questions:

1. How high is the building?

   Solving the first question is probably easy for you. The launch occurs at time , the initial value of the function. When , . The building is 60 feet high.

2. How high did the balloon travel?

   You answer the second question by finding the vertex of the parabola: . This gives you t, the number of seconds it takes the balloon to get to its highest point — 12 seconds after launch. Substitute the answer into the equation to get the height: . The balloon went 348 feet into the air.

3. If the statue is 10 feet tall, how many seconds did it take for the balloon to reach the statue after the launch?

   To solve the third question, use the fact that the statue is 10 feet high; you want to know when . Replace H with 10 in the equation and solve for t by factoring (see Chapters 1 and 3):

   

   When , . And when , .

   According to the equation, the amount of time is either 25 seconds or second. The doesn’t really make any sense because you can’t go back in time; if the balloon had started at the ground level, it would have taken that one second to reach that initial 60 feet in the air. The 25 seconds, however, tells you how long it took the balloon to reach the statue. Imagine the anticipation!