10.2: Solving Quadratic Equations by Graphing
Isaac Newton’s theory for projectile motion is represented by the equation:
\begin{align*}h(t)= \frac{1}{2} (g) t^2+v_0 t+h_0\end{align*}

\begin{align*}t=\end{align*}
t= time (usually in seconds) 
\begin{align*}g=\end{align*}
g= gravity due to acceleration; either \begin{align*}9.8 m/s^2\end{align*}9.8m/s2 or \begin{align*}32 ft/s^2\end{align*}32ft/s2 
\begin{align*}v_0=\end{align*}
v0= initial velocity 
\begin{align*}h_0=\end{align*}
h0= initial height of object
Consider the following situation: “A quarterback throws a football at an initial height of 5.5 feet with an initial velocity of 35 feet per second.”
By substituting the appropriate information:

\begin{align*}g=32\end{align*}
g=32 because the information is given in feet 
\begin{align*}v_0=35\end{align*}
v0=35 
\begin{align*}h_0=5.5\end{align*}
h0=5.5
The equation becomes \begin{align*}h(t)=\frac{1}{2} (32) t^2+35t+5.5 \rightarrow h(t)=16t^2+35t+5.5\end{align*}
Using the concepts from the previous lesson, we know
 The value of \begin{align*}a\end{align*}
a is negative, so the parabola opens downward.  The vertex is a maximum point.
 The \begin{align*}y\end{align*}
y− intercept is (0, 5.5).  The graph is narrow about its line of symmetry.
At what time will the football be 6 feet high? This equation can be solved by graphing the quadratic equation.
Solving a Quadratic Using a Calculator
Chapter 7 focused on how to solve systems by graphing. You can think of this situation as a system: \begin{align*}\begin{cases} y=16t^2+35t+5.5\\ y=6 \end{cases}\end{align*}. You are looking for the appropriate \begin{align*}x\end{align*}coordinates that give a \begin{align*}y\end{align*}coordinate of 6 feet. Therefore, you are looking for the intersection of the two equations.
Begin by typing the equations into the \begin{align*}[Y=]\end{align*} menu of your calculator. Adjust the window until you see the vertex, \begin{align*}y\end{align*}intercept, \begin{align*}x\end{align*}intercepts, and the horizontal line of 6 units.
By looking at the graph, you can see there are two points of intersection. Using the methods from chapter 7, find both points of intersection.
\begin{align*}(0.014,6) \ and \ (2.172,6)\end{align*}
At 0.014 seconds and again at 2.17 seconds, the football is six feet from the ground.
Using a Calculator to Find the Vertex
You can also use a graphing calculator to determine the vertex of the parabola. The vertex of this equation is a maximum point, so in the [CALCULATE] menu of the graphing option, look for [MAXIMUM].
Choose option #4. The calculator will ask you, “LEFT BOUND?” Move the cursor to the left of the vertex and hit [ENTER].
The calculator will ask, “RIGHT BOUND?” Move the cursor to the right of the vertex and hit [ENTER].
Hit [ENTER] to guess.
The maximum point on this parabola is (1.09, 24.64).
Example 1: Will the football reach 25 feet high?
Solution: The vertex represents the maximum point of this quadratic equation. Since its height is 24.64 feet, we can safely say the football will not reach 25 feet.
Example 2: When will the football hit the ground, assuming no one will catch it?
Solution: We know want to know at what time the height is zero. \begin{align*}\begin{cases} y=16t^2+35t+5.5 \\ y=0 \end{cases}\end{align*}. By repeating the process above and finding the intersection of the two lines, the solution is (2.33, 0). At 2.33 seconds, the ball will hit the ground.
The point at which the ball reaches the ground \begin{align*}(y=0)\end{align*} represents the \begin{align*}x\end{align*}intercept of the graph.
The \begin{align*}x\end{align*}intercept of a quadratic equation is also called a root, solution, or zero.
Example: Determine the number of solutions to \begin{align*}y=x^2+4\end{align*}.
Solution: Graph this quadratic equation, either by hand or with a graphing calculator. Adjust the calculator’s window to see both halves of the parabola, the vertex, the \begin{align*}x\end{align*}axis, and the \begin{align*}y\end{align*}intercept.
The solutions to a quadratic equation are also known as its \begin{align*}x\end{align*}intercepts. This parabola does not cross the \begin{align*}x\end{align*}axis. Therefore, this quadratic equation has no real solutions.
Example: Andrew has 100 feet of fence to enclose a rectangular tomato patch. He wants to find the dimensions of the rectangle that encloses the most area.
Solution: The perimeter of a rectangle is the sum of all four sides. Let \begin{align*}w=\end{align*} width and \begin{align*}l=\end{align*} length. The perimeter of the tomato patch is \begin{align*}100=l+l+w+w \rightarrow 100=2l+2w\end{align*}.
The area of a rectangle is found by the formula \begin{align*}A=l(w)\end{align*}. We are looking for the intersection between the area and perimeter of the rectangular tomato patch. This is a system.
\begin{align*}\begin{cases} 100 = 2l+2w \\ A = l(w) \end{cases}\end{align*}
Before we can graph this system, we need to rewrite the first equation for either \begin{align*}l\end{align*} or \begin{align*}w\end{align*}. We will then use the Substitution Property.
\begin{align*}100 &= 2l+2w \rightarrow 1002l=2w\\ \frac{1002l}{2} &= w \rightarrow 50l=w\end{align*}
Use the Substitution Property to replace the variable \begin{align*}w\end{align*} in the second equation with the express \begin{align*}50l\end{align*}.
\begin{align*}A=l(50l)=50ll^2\end{align*}
Graph this equation to visualize it.
The parabola opens downward so the vertex is a maximum. The maximum value is (25, 625). The length of the tomato patch should be 25 feet long to achieve a maximum area of 625 square feet.
Practice Set
Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both.
CK12 Basic Algebra: Solving Quadratic Equations by Graphing (10:51)
 What are the alternate names for the solution to a parabola?
 Define the following variables in the function \begin{align*}h(t)=\frac{1}{2} (g) t^2+v_0 t+h_0\end{align*}.
 \begin{align*}h_0\end{align*}
 \begin{align*}t\end{align*}
 \begin{align*}v_0\end{align*}
 \begin{align*}g\end{align*}
 \begin{align*}h(t)\end{align*}
 A rocket is launched from a height of 3 meters with an initial velocity of 15 meters per second.
 Model the situation with a quadratic equation.
 What is the maximum height of the rocket? When will this occur?
 What is the height of the rocket after four seconds? What does this mean?
 When will the rocket hit the ground?
 At what time will the rocket be 13 meters from the ground?
How many solutions does the quadratic equation have?
 \begin{align*}x^2+3=0\end{align*}
 \begin{align*}2x^2+5x7=0\end{align*}
 \begin{align*}x^2+x3=0\end{align*}
Find the zeros of the quadratic equations below. If necessary, round your answers to the nearest hundredth.
 \begin{align*}y=x^2+4x4\end{align*}
 \begin{align*}y=3x^25x\end{align*}
 \begin{align*}x^2+3x+6=0\end{align*}
 \begin{align*}2x^2+x+4=0\end{align*}
 \begin{align*}x^29=0\end{align*}
 \begin{align*}x^2+6x+9=0\end{align*}
 \begin{align*}10x^23x^2=0\end{align*}
 \begin{align*}\frac{1}{2}x^22x+3=0\end{align*}
 \begin{align*}y=3x^2+4x1\end{align*}
 \begin{align*}y=94x^2\end{align*}
 \begin{align*}y=x^2+7x+2\end{align*}
 \begin{align*}y=x^210x25\end{align*}
 \begin{align*}y=2x^23x\end{align*}
 \begin{align*}y=x^22x+5\end{align*}
 Andrew is an avid archer. He launches an arrow that takes a parabolic path, modeled by the equation \begin{align*}y=4.9t^2+48t\end{align*}. Find how long it takes the arrow to come back to the ground.
For questions 24 – 26,
(a) Find the roots of the quadratic polynomial.
(b) Find the vertex of the quadratic polynomial.
 \begin{align*}y=x^2+12x+5\end{align*}
 \begin{align*}y=x^2+3x+6\end{align*}
 \begin{align*}y=x^23x+9\end{align*}
 Sharon needs to create a fence for her new puppy. She purchased 40 feet of fencing to enclose three sides of a fence. What dimensions will produce the greatest area for her puppy to play?
 An object is dropped from the top of a 100foottall building.
 Write an equation to model this situation.
 What is the height of the object after 1 second?
 What is the maximum height of the object?
 At what time will the object be 50 feet from the ground?
 When will the object hit the ground?
Mixed Review
 Factor \begin{align*}3r^24r+1\end{align*}.
 Simplify \begin{align*}(2+\sqrt{3})(4+\sqrt{3})\end{align*}.
 Write the equation in slopeintercept form and identify the slope and \begin{align*}y\end{align*}intercept: \begin{align*}93x+18y=0\end{align*}.
 The half life of a particular substance is 16 days. An organism has 100% of the substance on day zero. What is the percentage remaining after 44 days?
 Multiply and write your answer in scientific notation: \begin{align*}0.00000009865 \times 123564.21\end{align*}
 A mixture of 12% chlorine is mixed with a second mixture containing 30% chlorine. How much of the 12% mixture is needed to mix with 80 mL to make a final solution of 150 mL with a 20% chlorine concentration?
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