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word problem involving a quadratic function

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Let's take this problem for an example. (I'll state the whole problem first, then go about showing how to solve it. It's long, but it covers about every special case I can think of. I use this sort of problem in my college classes to have an example for all the different nuances of quadratic function modelling - word problems.)

PROBLEM:

A ball is thrown into the air. Its height above the ground, h (measured in feet), at any given time after the ball is thrown, t (measured in seconds), can be modelled using the quadratic function h(t) = -16t2 + 16t + 32.

a) From what height above the ground was the ball thrown?

b) How many seconds after the ball was thrown did it reach its maximum height?

c) What was the ball's maximum height?

d) When did the ball hit the ground?

 

SOLUTION:

a) In all of these problems, we are given one piece of information and asked to find the other. However, the piece of information we are given is not always given to us in a very obvious manner. Take, for instance, this question. Between height and time, which piece of information are we given. If we pay close attention, it asks us 'From what height...' which means we weren't given height; we must have been given time. So now, we look at the problem to find any information related to time. (This will involve reading the overall prompt and the question in part (a) again.)

In the overall prompt, we are told that t represents any amount of time after the ball is thrown, so if we are talking about when the ball is being thrown, what time value do we have? No time has elapsed yet because the ball hasn't been thrown, so the moment the ball is thrown is represented by t = 0. If we substitute this known piece of information into our function, we get

h(0) = -16(0)2 + 16(0) + 32 = 0 + 0 + 32 = 32

So the ball was thrown from a height of 32 ft above the ground.

 

b) This part of the problem requires us to recognize that a quadratic function has the graph of a parabola. The vertex of a parabola is the point on the graph of the function which has a unique function value - that is, it doesn't have a matching function value 'on the other side' of the parabola; it is the tip of the parabola. In this problem, the ball is only going to reach this maximum height once. The ball will reach every other height twice - once going up and again going down - but not its maximum height. So, in this problem, the maximum height represents the vertex of the parabola.

In order to algebraically find the vertex of a parabola, we can use the formula t = -b/(2a) - where t represents the independent variable of the function (in this case, time), b represents the coefficient on the single-power term of the independent variable (in this case, the coefficient of the middle term, 16), and a represents the leading coefficient - that is, the coefficient on the double-power term of the independent variable (in this case, -16).

So, for this part of the problem, where we are asked only about the time related to the maximum height of the ball, we substitute for b and a and calculate to find t.

t = -(16) / (2 * (-16)) = -16 / -32 = 1/2 = 0.5 seconds

So the ball reaches its maximum height one-half second after it is thrown into the air.

 

c) For this part of the problem, we can use the piece of information we found in part (b) - that the ball reached its maximum height 0.5 seconds after it was thrown - to find what that height was. We simply substitute t = 0.5 into the h(t) function.

h(0.5) = -16(0.5)2 + 16(0.5) + 32

h(0.5) = -16(0.25) + 8 + 32

h(0.5) = -4 + 8 + 32 = 4 + 32 = 36

So the maximum height the ball reached was 36 feet above the ground.

 

d) The last part of this problem asks us to find when the ball hit the ground. What piece of information are we looking for, and what piece of information do we already have? The question asks "When...". Does that refer to time or height? Time. So we know we are looking for - and therefore don't already have - a value for time. That means that we must have a value for height. What is that value.

The problem tells us that we are looking for the time when the ball has hit the ground. We know that h represents the height above the ground, in feet, so what value is the ground represented by? The ground, being 0 feet above the ground, is represented by the h = 0. Since we have one piece of information and we are looking for the other, we can use our function to find it.

However, we will use it differently than in part (a). In part (a), we were given time and were looking for height, so we put our known value into the function for t. In this part of the problem, we are given height and are looking for time, so we will put our known value into the function for h - or h(t) - and solve for t.

(0) = -16t2 + 16t + 32

In order to solve this equation for t, we must factor. In this case, before we factor into the product of two binomials, we can factor out a greatest common factor (GCF) of 16 - more specifically, -16, which will change all the signs. (We do this because it is easier to fact a trinomial - a three-term polynomial expression - into the product of two binomial expressions if the leading coefficient is +1.) When we factor out the -16, we are left with:

0 = -16 (t2 - t - 2)

We can divide both sides by -16 now so that we get:

0 / -16 = -16 / -16 * (t2 - t - 2)

0 = (t2 - t - 2)

To factor this trinomial, we look for two numbers which multiply to be -2 and which add (or subtract) to be -1 (the coefficient on the single-term power of the variable). In this case, -2 and +1 work.

-2 * 1 = -2

-2 + 1 = -1

Therefore, our quadratic equation becomes:

0 = (t - 2) * (t + 1)

Using the Zero Product Property, if two things multiplied together equals zero, one or both of them is equal to zero. Since the product of our two binomial expressions is equal to 0, we take both binomial expressions and set each of them equal to zero individually.

t - 2 = 0    => t = 2

t + 1 = 0   => t = -1

Because t represents the number of seconds after the ball was thrown, a negative number for t represents the number of seconds before the ball was thrown. We are not concerned about these values because they will not be accurate in the context of our problem. (That is why math can sometimes only model a real-world problem, as in this case. A part of the model works in the context of our problem, and another part doesn't. We ignore the part that doesn't, so in this case, we ignore negative values for t.)

The only real option out of the values we have is t = 2. Because, at the beginning of this part of the problem, we set h = 0 to find this value, the ball hit the ground 2 seconds after it was thrown.