This section covers:
 Quadratic Projectile Problem
 Quadratic Trajectory (Path) Problem
 Optimization of Area Problem
 Maximum Profit and Revenue Problems
 Population Problem
 Linear Increase/Decrease Problem
 Pythagorean Theorem Quadratic Application
 Quadratic Inequality Problem
 Finding Quadratic Equation from Points or a Graph
Quadratic applications are very helpful in solving several types of word problems, especially where optimization is involved. Again, we can use the vertex to find the maximum or the minimum values, and roots to find solutions to quadratics.
Note that we did a Quadratic Inequality Real World Example here. Note also that we will discuss Optimization Problems using Calculus in the Optimization section here.
Quadratic Projectile Problem:
Quadratic Projectile problems are common quadratic application problems.
Problem:
Jennifer hit a golf ball from the ground and it followed the projectile \(h\left( t \right)=16{{t}^{2}}+100t\), where \(t\) is the time in seconds, and \(h\) is the height of the ball. Find the highest point that her golf ball reached and also when it hits the ground again. Find a reasonable domain and range for this situation.
Solution: Note that in this example, we are using the generic equation \(h\left( t \right)=16{{t}^{2}}+{{v}_{0}}t+{{h}_{0}}\), where, in simplistic terms, the –16 is the gravity (in feet per seconds per seconds), the \({{v}_{0}}\) is the initial velocity (in feet per seconds) and the \({{h}_{0}}\) is the initial height (in feet). (If units are in meters, the gravity is –4.9 meters per second per second). Since we need to find the highest point of the ball, we need to get the vertex of the parabola. We also need to know when the height \(h\) is back to 0 again. Then we can use these two values to find a reasonable domain and range. Let’s get the vertex both algebraically and using a Graphing Calculator: 

Solve Quadratic Algebraically  Solve Quadratic with Graphing Calculator 
\(h\left( t \right)=16{{t}^{2}}+100t\)
To get the vertex, we can use (\(\displaystyle \frac{b}{{2a}}\), plug \(\displaystyle \frac{b}{{2a}}\) into the \(t\) to get the \(y\)) to find the coordinates of the vertex, when \(y=a{{t}^{2}}+bt+c\):
\(\displaystyle \frac{b}{{2a}}=\frac{{100}}{{32}}=\,\,\,\,3.125\, \text{seconds}\)
(This is the time, and to get the height, we plug this into \(16{{t}^{2}}+100t\), and get 156.25 feet.)
To get when the ball hits the ground, we have to set \(16{{t}^{2}}+100t\) to 0; we get \(t=6.25\) seconds. This is the second root.
This makes sense, since the ball started from the ground, so the parabola is symmetrical around the line of symmetry, which is \(x=3.125\). 
Note that I used ZOOM 6, ZOOM 0, and ZOOM 3 ENTER a few times so I could see the vertex in the window.
Then I used 2^{nd} TRACE (CALC), 4 (maximum), moved the cursor to the left of the top after “Left Bound?”, hit ENTER, moved the cursor to the right of the top after “Right Bound?”, and then hit ENTER twice to get the vertex. We see that the \(y\)value of the vertex, the height, is 156.25 feet.
To get the root, push 2^{nd} TRACE (CALC), and then push 2 for ZERO (or move cursor down to ZERO). The calculator will then say “Left Bound?” Using the cursors, move the cursor anywhere to the left of the zero (where the graph hits the \(x\)axis) and hit ENTER. We want the zero that is positive. When the calculator says “Right Bound?” move the cursor anywhere to the right of that zero and hit ENTER. The calculator will say “Guess?”. Hit ENTER once more, and you have your zero, which is 6.25 feet. 
The reasonable domain is \(\left[ {0,6.25} \right]\) and the reasonable range is \(\left[ {0,156.25} \right]\).

(We will discuss projectile motion using parametric equations here in the Parametric Equations section.)
Note that the independent variable represents time, not distance; sometimes parabolas represent the distance on the \(x\)axis and the height on the \(y\)axis, and the shapes are similar. Height versus distance would be the path or trajectory of the bouquet, as in the following problem.
Quadratics Trajectory (Path) Problem
Note that in this problem, the \(x\)axis is measuring the horizontal distance of the path of the ball, not the time, so when we draw the parabola, it’s a true indication of the trajectory or path of the ball.
Quadratic Application Problem 
Solution 
Audrey throws a ball in the air, and the path the ball makes is modeled by the parabola \(y8=0.018{{\left( {x20} \right)}^{2}}\), measured in feet.
What is the maximum height the ball reaches, and how far (horizontally) from Audrey does is the ball at its maximum height?
How far does the ball travel before it hits the ground? 
Since the quadratic is basically already in vertex form (\(y=a{{\left( {xh} \right)}^{2}}+k\), where \((h,k)\) is the vertex), we can see that the vertex from \(0=0.018{{\left( {x20} \right)}^{2}}+8\) is \((20,8)\).
This means that the maximum height (since the parabola opens downward) is 8 feet and it happens 20 feet away from Audrey.
When the ball hits the ground, \(y=0\), so we have \(0=0.018{{\left( {x20} \right)}^{2}}+8\). We could expand the binomial and use the quadratic formula, but it’s much easier to use the square root method, since we have a square in the original: \(\begin{align}0&=0.018{{\left( {x20} \right)}^{2}}+8\\\frac{{8}}{{0.018}}&={{\left( {x20} \right)}^{2}}\\\pm \sqrt{{\frac{8}{{0.018}}}}&=\sqrt{{{{{\left( {x20} \right)}}^{2}}}}\\x20&=\pm 21.08\\x&=41.08\end{align}\) Note that we had to “throw away” the negative part of the solution. The ball will hit the ground 41.08 feet from Audrey. (We could have also used a graphing calculator to solve this problem.) 
A kicker kicks a football in an attempt to make a field goal from 105 feet away from the uprights. The uprights are 14 feet high.
The football reaches its maximum height of 75 feet when it is 50 feet from the kicker.
What is the height of the ball when it reaches the uprights? Did the kicker make a field goal (did the ball go over the uprights?) 
First sketch the problem; we want the height of the ball (\(y\)) when it reaches the uprights :
Get the equation of the quadratic for the path (height) of the ball, given the distance from the kicker. Since the vertex is at \(\left( {50,75} \right)\), we have \(y=a{{\left( {x50} \right)}^{2}}+75\) (we know \(a\) will be negative). We can use the fact that \(\left( {0,0} \right)\) is a point to get \(a\): \(\begin{align}0&=a{{\left( {050} \right)}^{2}}+75\\75&=2500a;\,\,\,\,a=\frac{3}{{100}}\\y&=\frac{3}{{100}}{{\left( {x50} \right)}^{2}}+75\end{align}\) Now get the height of the ball when the distance from the kicker (\(x\)) is 105 feet: \(\displaystyle y=\frac{3}{{100}}{{\left( {10550} \right)}^{2}}+75=14.25\,\text{feet}>14\,\text{feet}\). Yes! It barely goes over the upright, so the kicker made a goal! 
Optimization of Area Problem:
Here’s a very common application of quadratics called optimization. Optimization typically involves finding the vertex of a parabola since it’s the highest or lowest amount.
Maximum Profit Problem:
Here’s another optimization problem:
Quadratic Maximum Profit Problem 
Solution 
The profit from selling local ballet tickets depends on the ticket price. Using past receipts, the profit can be modeled by the function \(p=15{{x}^{2}}+600x+60\), where \(x\) is the price of each ticket.
What is the ticket price that gives the maximum profit, and what is that maximum profit? 
This problem is actually much easier since we are given the formula for the profit, given the price of each ticket.
Simply either graph the function to get the vertex, or use (\(\displaystyle \frac{b}{{2a}}\), plug \(\displaystyle \frac{b}{{2a}}\) into the \(x\) to get the \(y\)) to find the coordinates of the vertex:
\(\displaystyle \frac{b}{{2a}}=\frac{{600}}{{2\left( {15} \right)}}=20,\,\,\,\,\,\,f\left( {20} \right)=15{{\left( {20} \right)}^{2}}+600\left( {20} \right)+60=6060\)
Since the vertex is \(\left( {20,6060} \right)\), the ticket price should be $20 to maximize profit, and that maximum profit is $6060. 
Maximum Revenue Problem:
Here’s another optimization problem:
Bunny Rabbit Population Problem:
Quadratic Application Problem  Solution and Calculator Instructions 
The observed bunny rabbit population on an island is given by the function \(p=.4{{t}^{2}}+130t+1200\), where \(t\) is the time in months since they began observing the rabbits.
(a) When is the maximum population attained?
(b) What is the maximum population?
(c) When does the bunny rabbit population disappear from the island?

Find the vertex on the graphing calculator the same way we did above using the 2nd trace (Calc) Maximum feature. We see that the vertex is \(\left( {162.5,11762.5} \right)\). The maximum rabbit population was roughly 11762 rabbits (we can’t have half of a rabbit!) when it was 162.5 months after they began observing the rabbit population. This answers (a) and (b).
For (c), we need to see when the graph goes back down to 0; this is when there are no rabbits left on the island. To get the roots, push 2^{nd} TRACE (CALC), and then push 2 for ZERO (or move cursor down to ZERO). The calculator will then say “Left Bound?” Using the cursors, move the cursor anywhere to the left of the zero (where the graph hits the \(x\)axis) and hit ENTER. We want the zero that is positive. When the calculator says “Right Bound?” move the cursor anywhere to the right of that zero and hit ENTER. The calculator will say “Guess?”. Hit ENTER once more, and you have your zero. Note that sometimes the calculator gets confused and gives you a number way close to 0 (1E9) instead of 0 for the \(y\)value. So, for answer (c) , the rabbit population will disappear from the island at around 334 months from when the observations started.
Note that you can also find the roots by setting “\({{Y}_{2}}=\) ” to 0 and use the Intercept function to find the roots. You may have to use the ZOOM and/or WINDOWS to make sure you see the point of intersection you want. You may also have to use TRACE and then arrows to move the cursor close to the point of intersection you want (the positive one), if the cursor is closer to the other root. To get the point(s) of intersection, push 2^{nd} TRACE (CALC), and then either push 5, or move cursor down to intersect. You should see “First curve?” at the bottom. Then push ENTER, ENTER, ENTER. 
Linear Increase/Decrease Problem:
OK, use your imaginations on this one (sorry!):
Problem:
Taylor and Miranda are performing on a magic dimensionchanging stage that is 20 yards long by 15 yards wide. The length is decreasing linearly (with time) at a rate of 2 yards per hour, and the width is increasing linearly (with time) at a rate of 3 yards per hour. When will the stage have the maximum area, and when will the stage disappear (has an area of 0 square yards)? Solution: This one’s a little trickier, since we are asking when the stage will be the greatest area, and when it will have an area of 0, yet we are only given distances and rates. Since we’re finding areas, we need to work with distances only. And we know that \(\text{Distance}=\text{Rate}\times \text{Time}\). Do you see how at time \(t\), the length of the stage is \((202t)\) and the width is \((15+3t)\)? Think about it: after one hour, the length of the stage will have decreased by 2 yards, and the width will have increased by 3 yards, so the new stage will be 18 by 18 yards. After two hours, the length will be 16 yards, and the width will be 21 yards, and so on. Now, let’s find the answers, with and without using the graphing calculator: 

Solve Quadratic Algebraically  Solve Quadratic with Graphing Calculator 
\(A\left( t \right)=\left( {202t} \right)\left( {15+3t} \right)\)
To get the vertex, we can use \(\displaystyle \frac{b}{{2a}}\) (\(\displaystyle \frac{b}{{2a}}\), plug into the \(t\) to get the \(A(t)\) or \(y\)) to find the coordinates of the vertex, when \(y=a{{t}^{2}}+bt+c\):
\(y=\left( {202t} \right)\left( {15+3t} \right)\,\,\,\,\text{or}\,\,\,y=6{{t}^{2}}+30t+300\) \(\displaystyle \,\,\frac{{b}}{{2a}}=\frac{{30}}{{12}}=\,2.5 \text{ hours}\)
This is the time that the area of the stage will be at a maximum.
Now plug in 2.5 (\(t\)) to get the \(y\), or the area:
\(\begin{align}A\left( t \right)&=6{{t}^{2}}+30t+300\\&=6{{\left( {2.5} \right)}^{2}}+30\left( {2.5} \right)+300\\&=337.5 \text{ yard}{{\text{s}}^{\text{2}}}\end{align}\)

Note that I used ZOOM 6, ZOOM 0, and ZOOM 3 ENTER a few times so I could see the vertex in the window.
Then I used 2^{nd} TRACE (CALC), 4 (maximum), moved the cursor to the left of the top after “Left Bound?”, moved the cursor to the right of the top after “Right Bound?”, and then hit ENTER twice to get the vertex.
Since this is the \(t\) part of the vertex, 2.5 hours is the time that maximizes the area.
The area is \(y\) part of the vertex, which is \(337.5\) yards^{2}. 
Now we need to find when the stage will have no area left. We need to set the equation to 0, or find the rightmost root with the calculator:  
\(0=\left( {202t} \right)\left( {15+3t} \right)\)
Since this is already in factored form, we can just set each factor to 0 to see when the quadratic will be 0:
\(\displaystyle \begin{array}{l}202t=0\,\,\,\,\,\,15+3t=0\\\,\,\,\,20=2t\,\,\,\,\,\,\,\,\,15=3t\,\,\,\\\,\,\,\,\,t=10\,\,\,\,\,\,\,\,\,\,\,\,t=5\end{array}\)
Since we can’t have a negative time, we have to “throw away” the –5 root.
The time that the area will be 0 is in 10 hours. 
To get the roots, push “2^{nd} TRACE” (CALC), and then push 2 for ZERO (or move cursor down to ZERO).
The calculator will then say “Left Bound?” Using the cursors, move the cursor anywhere to the left of the zero (where the graph hits the \(x\)axis) and hit ENTER. We want the zero that is positive (rightmost zero). When the calculator says “Right Bound?” move the cursor anywhere to the right of that zero and hit ENTER.
The calculator will say “Guess?”. Hit ENTER once more, and you have your zero. We see that \(x=10\); the time the area will be zero is in 10 hours.

Pythagorean Theorem Quadratic Application:
OK, here’s one where you’ll use a bit of Geometry. You probably learned the Pythagorean Theorem awhile back – it’s the one with the right triangle and all the squares in it!
Here is the type of problem you may get:
Quadratic Problem and Picture  Solution 
The hypotenuse of a right triangle is 4 inches longer than one leg and 2 inches longer than the other. Find the dimensions of the triangle.
Also, find a reasonable domain for the hypotenuse.
I decided to make the hypotenuse of the triangle the \(x\)value since it was easier to get a reasonable domain for the hypotenuse: To get the reasonable domain for the hypotenuse, we know it has to be greater than 0, and since we have minus signs in the expressions for the legs, we have to look at those, too. Both of the legs must have values that are positive, so “\(x2\)”, and “\(x4\)” both must be positive. \(x\) has to be greater than 4 (do you see why?). Therefore, the reasonable domain for the hypotenuse is \(x>4\), or \(\left( {4,\infty } \right)\). 
We can use the Pythagorean Theorem to set up the problem. Remember that the sum of the square of the two legs (the sides next to the right angle) add up to the square of the hypotenuse: \({{\left( {x2} \right)}^{2}}+{{\left( {x4} \right)}^{2}}={{x}^{2}}\)
We can then multiply out (FOIL) the binomials and put everything to one side and factor to get the answers: \(\begin{array}{c}\,{{\left( {x2} \right)}^{2}}+{{\left( {x4} \right)}^{2}}={{x}^{2}}\\{{x}^{2}}4x+4+{{x}^{2}}8x+16={{x}^{2}}\\{{x}^{2}}12x+20=0\\\left( {x10} \right)\left( {x2} \right)=0\\x=2,\,\,10\end{array}\)
We have to “throw away” the value of 2, since if we plugged it in to get the sides, we’d have negative numbers (sides can’t be negative!) (This value isn’t in the domain.)
The \(x\)value is 10 inches (hypotenuse), and the sides are 8 inches and 6 inches.
Let’s check our answers: \({{8}^{2}}+{{6}^{2}}={{10}^{2}}\). 
Quadratic Inequality Problem:
You may encounter a problem like this – which is really not too difficult.
Finding Quadratic Equations from Points or a Graph
We saw how to find a Quadratic Equation from a point and/or graph here in the Solving Quadratics by Factoring and Completing the Square section. Here is one more problem.
Solution:
Quadratic Problem and Graph 
Solution 
Emmy throws a dog toy up in the air from 5 feet above the ground. When the toy is 2 feet from the her, the toy reaches a maximum height of 9 feet, and then lands back on the ground 5 feet from her.
Find the “\(a\)” (the coefficient of the \({{x}^{2}}\)) for the parabola of the flight of the toy, and write this quadratic equation in vertex form, standard form, and factored form.

We know the vertex is \((2,9)\), and the \(y\)intercept is \((0,5)\). “\(a\)” is negative, since the parabola faces downwards. Putting the equation in vertex form would be \(y=a{{\left( {x2} \right)}^{2}}+9\).
Let’s get the “\(a\)”. Since we only have the vertex form at this point, we can’t use the vertex as the point to plug in (try it and you’ll see why!). Use the \(y\)intercept \((0,5)\) to plug in to \((x,y)\) to get “\(a\)”: \(\begin{align}y&=a{{\left( {x2} \right)}^{2}}+9\\5&=a{{\left( {02} \right)}^{2}}+9\\5&=4a+9\\4&=4a;\,\,\,\,\,a=1\end{align}\)
The vertex form is \(y=1{{\left( {x2} \right)}^{2}}+9\) (or \(y={{\left( {x2} \right)}^{2}}+9\)), and the standard form (by multiplying it out) is \(y={{x}^{2}}+4x+5\).
We can factor this to get the factored form, which is \(y=\left( {x5} \right)\left( {x+1} \right)\) (take the negative out first). You can check all three forms by putting them all in a <strong “>graphing calculator (like in \(\displaystyle {{Y}_{1}},\,\,\,{{Y}_{2}},\,\,\,{{Y}_{3}}\)) to make sure they are all the same parabola! (You can see from the graph that this looks correct; the roots look like \((0,1)\) and \((5,0)\).) 
Learn these rules, practice, practice, practice, and you’ll rock at math!
On to Solving Absolute Value Equations and Inequalities – you’re ready!