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Precalculus MATH F156X Summer 2026

Dr. Latrice Bowman

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Section GR-1 Quadratic Functions and Models

Worksheet Graphing Quadratics and Quadratic Models

For this outcome Specifically, for this outcome, students should be able to:
  • Find the intercepts of a quadratic function.
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  • Find the vertex of a quadratic function.
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  • Find the axis of symmetry of a quadratic function.
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  • Rewrite a quadratic function in vertex form.
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  • Use a quadratic function to model real-world situations.
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Quadratic Functions
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A quadratic function is a function of the form \(f(x) = ax^2 + bx + c\text{,}\) where \(a\text{,}\) \(b\text{,}\) and \(c\) are constants and \(a \ne 0\text{.}\)
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Vertex Form of a Quadratic Function
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The vertex form of a quadratic function is given by \(f(x) = a(x - h)^2 + k\text{,}\) where \((h, k)\) is the vertex of the parabola.
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One can convert a quadratic function from standard form to vertex form by completing the square. To complete the square, we first need to factor out the coefficient of \(x^2\) from the first two terms of the function. We then take the coefficient of \(x\text{,}\) divide it by 2, and square it. We then add and subtract this value inside the function. The last step is to factor the perfect square trinomial.
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The vertex of a parabola can also be found using the formula \((h, k) = \left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right)\text{.}\)
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Graphing Quadratics
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To graph a quadratic function, we can use the vertex form of the function. The vertex of the parabola is given by \((h, k)\text{.}\) The axis of symmetry is the vertical line \(x = h\text{.}\) The direction of opening is determined by the sign of \(a\text{.}\) If \(a \gt 0\text{,}\) the parabola opens upward; if \(a \lt 0\text{,}\) it opens downward.
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Modeling Quadratic Functions
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Modeling quadratic functions involves using quadratic equations to represent real-world situations. For example, the height of a projectile over time can be modeled by a quadratic function. The vertex of the parabola represents the maximum or minimum value of the function, depending on whether the parabola opens upward or downward.
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To model a situation with a quadratic function, we need to identify the key features of the situation, such as the vertex, axis of symmetry, and direction of opening. We can then use these features to write the equation of the quadratic function that models the situation.
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Strategy for solving word problems:
  1. Read and re-read the problem carefully.
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  2. Identify what is being asked and what information is given.
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  3. Define variables to represent the unknown quantities.
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  4. Draw a diagram or use a table to organize the information.
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  5. Determine if there is a formula that relates the known and unknown quantities.
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  6. Form an equation based on the information and the formula.
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  7. Solve the equation.
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  8. Check the solution and interpret it in the context of the problem.
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Worksheet GR-1 Quadratic Functions and Models

Quadratic Functions
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Quadratic Models
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Section GR-1 Rubric

Worksheet Worksheet

Before attempting the GR-1 outcome, you should review the following rubric. This shows the criteria you will be graded on and the expectations to earn each mastery level.
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Table 14. GR1 Rubric
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Criteria M P R N
Simplifying Algebraic Expressions All answers are fully simplified. Exact answers are given, not approximations. Radical and exponential expressions are used appropriately and are simplified in the correct manner No algebraic errors OR some expressions are not simplified. There are some algebraic errors and some expressions were not simplified. There are some algebraic errors and some expressions were not simplified.
Basics of Graphing Graph is labeled with units for each axis. The graph includes all of the important features (intercepts, asymptotes, deleted points, etc).  The graph is appropriately drawn to accommodate all features. Graph is labeled with units for each axis and includes most of the important features or the graph has all of the important features but is not labeled. Graph is not labeled and has some of the important features. No graph was given or the given graph is not labeled and does not have the correct features of the function.
Problem Structure and Mathematical Communication Problems have a clear beginning, middle, and end. Work progresses clearly from one step to the next. The work provided uses appropriate methods and notation and provides a clear solution. Problems have clear beginnings and ends. Work progresses from one step to the next. The work provided uses algebraic methods, correct notation, and provides a solution. Parts of the mathematical structure are missing, or steps in the progress of the solution are missing. There are errors in the mathematical notation. The work provided uses algebraic methods and provides a solution. There is not a clear structure to the solution, notation is used incorrectly, and it is unclear what methods are being used.
Give the vertex form of a quadratic function Correctly apply completing the square to rewrite the function. The final answer is a function written in vertex form with correct notation. Correctly apply completing the square with minor errors to rewrite the function. The final answer is written in vertex form. Found the vertex of the quadratic function, but did not complete the square, or did not use the vertex form, or had multiple errors in the process. No vertex form of the function was given. Not enough evidence to show knowledge of completing the square.
Find the attributes of a quadratic function Found both the x and y intercepts with correct notation.  Found the range using correct interval notation. Found the x or the y intercepts with correct notation.  Found a range using interval notation. Both the x and y intercepts are incorrect and a range was given. No intercepts or range were given.
Graph a quadratic functions Graph contains all relevant information; has correct end-behavior, vertex, intercepts and slopes. Graph contains most of the relevant information; has partially correct end-behavior, slopes, vertex or intercepts. A graph was given but it does not represent a quadratic function. No graph is given.
Setup a quadratic model Identified and defined the domain variable; created a quadratic function in terms of the domain variable. Identified the practical domain with correct notation. Defined a domain variable; created a quadratic function in terms of the domain variable with minor errors. Gave a practical domain with minor errors. Identified a domain variable; created a function; gave a domain. Did not define or identify the domain variable; did not give a function; did not give a domain.
Solve a quadratic model Used quadratic methods to solve the model with a correct solution using correct notation. Used quadratic methods to solve the model with minor errors. Used algebraic methods to solve the model. No algebraic methods or quadratic methods were shown to get an answer.
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Worksheet GR-1 Quadratic Functions and Models

This outcome covers quadratic functions and models. You should be able to rewrite quadratic functions in vertex form. Be able to find the domain, range, and intercepts of the function. You should also be able to graph quadratic functions. Here you will also be asked to setup and solve real-world problems using quadratic models. You should be able to write out complete steps to solve these types of problems by hand, without the use of a calculator.
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For the quiz on this outcome you would be given 30 minutes to complete 2 problems. Here are samples of the types of problems you will encounter:
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1.

Given the function \(f(x)=\frac{1}{2}x^2+5x+8\text{.}\)
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(a)
Write the function in vertex form.
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Hint.
Use the method of completing the square to rewrite the function in vertex form.
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Answer.
\(f(x)=\frac{1}{2}(x+5)^2-\frac{9}{2}\)
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Solution.
Starting with \(f(x)=\frac{1}{2}x^2+5x+8\text{,}\) we factor out the coefficient of \(x^2\) from the first two terms:
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\(f(x)=\frac{1}{2}(x^2+10x)+8\)
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To complete the square, we take half of the coefficient of \(x\) and square it: \(\left(\frac{10}{2}\right)^2 = 5^2 = 25\text{.}\)
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We add this value inside the parentheses and subtract this value multiplied by the coefficient to the outside of the parentheses:
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\(f(x)=\frac{1}{2}(x^2+10x+25)-\frac{1}{2}(25)+8\)
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We can now simplify the perfect square trinomial and the constant term:
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\(f(x)=\frac{1}{2}[(x+5)^2]-\frac{25}{2}+\frac{16}{2}\)
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Simplifying:
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\(f(x)=\frac{1}{2}(x+5)^2-\frac{9}{2}\)
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(b)
State the intercepts.
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Hint.
To find the y-intercept, evaluate the function at \(x=0\text{.}\) To find the x-intercepts, set the function equal to zero and solve for \(x\text{.}\)
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Answer.
y-intercept: \((0,8)\text{,}\) x-intercepts: \((-2,0)\) and \((-8,0)\)
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Solution.
To find the y-intercept, we evaluate the function at \(x=0\text{:}\)
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\(f(0)=\frac{1}{2}(0)^2+5(0)+8=8\)
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Thus, the y-intercept is \((0,8)\text{.}\)
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To find the x-intercepts, we set the function equal to zero and solve for \(x\text{:}\)
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\(0=\frac{1}{2}x^2+5x+8\)
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Multiplying both sides by 2 to clear the fraction:
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\(0=x^2+10x+16\)
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We can now factor the quadratic:
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\(0=(x+5)^2-25+16\)
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\(0=(x+5)^2-9\)
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\((x+5)^2=9\)
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\(x+5=\pm 3\)
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\(x=-5\pm 3\)
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Thus, the x-intercepts are \((-5+3,0)\) and \((-5-3,0)\text{,}\) which can be written as \((-2,0)\) and \((-8,0)\text{.}\)
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(c)
State the range.
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Hint.
Look at the leading coefficient and the k-value of the vertex form.
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Answer.
\(\left[-\frac{9}{2},\infty\right)\)
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Solution.
Since the coefficient of \(x^2\) is positive, the parabola opens upwards. The vertex represents the minimum point of the function, so the range starts at the y-coordinate of the vertex and extends to infinity. This gives us the range \(\left[-\frac{9}{2},\infty\right)\text{.}\)
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(d)
Graph f(x).
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Hint.
Plot the vertex and the intercepts, then sketch the parabola.
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Answer.
Diagram Exploration Keyboard Controls
Key Action
Enter, A Activate keyboard driven exploration
B Activate menu driven exploration
Escape Leave exploration mode
Cursor down Explore next lower level
Cursor up Explore next upper level
Cursor right Explore next element on level
Cursor left Explore previous element on level
X Toggle expert mode
W Extra details if available
Space Repeat speech
M Activate step magnification
Comma Activate direct magnification
N Deactivate magnification
Z Toggle subtitles
C Cycle contrast settings
T Monochrome colours
L Toggle language (if available)
K Kill current sound
Y Stop sound output
O Start and stop sonification
P Repeat sonification output
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Solution.
To graph the function, we start by plotting the vertex, which is at \((-5,-\frac{9}{2})\text{.}\) Next, we plot the y-intercept at \((0,8)\) and the x-intercepts at \((-2,0)\) and \((-8,0)\text{.}\) Finally, we sketch the parabola opening upwards through these points.
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2.

A local theatre group is selling tickets for a play to raise money for new equipment. They find that if they charge $10 per ticket, they will sell 200 tickets. For every $1 increase in price, they will sell 10 fewer tickets. What price should they charge and how many tickets should they sell to maximize their revenue?
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(a)
Define the variable for your function.
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Hint.
Pick a variable and state what it represents in the context of the problem.
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Answer.
Let \(x\) represent the number of $1 increases in price.
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(b)
Set up the quadratic model.
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Hint.
Write the revenue function in terms of the variable you defined. Remember that revenue is price multiplied by quantity.
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Answer.
\(R(x)=(10+x)(200-10x)\) or \(R(x)=-10x^2+100x+2000\)
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Solution.
To set up the revenue function, we first express the price and quantity in terms of \(x\text{.}\) The price can be expressed as \(10+x\text{,}\) since the base price is $10 and it increases by $1 for each increase. The quantity can be expressed as \(200-10x\text{,}\) since they sell 200 tickets at the base price and sell 10 fewer tickets for each increase. Thus, the revenue function is: \(R(x)=(10+x)(200-10x)\) or \(R(x)=-10x^2+100x+2000\)
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(c)
State the practical domain.
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Hint.
Consider the context of the problem to determine the practical domain.
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Answer.
\([0,20]\)
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Solution.
Since \(x\) represents the number of $1 increases in price, it cannot be negative, so the lower bound of the domain is 0. The upper bound is 20, since if they increase the price by $20, they will not sell any tickets. Thus, the practical domain is \([0,20]\text{.}\)
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(d)
Find the price and number of tickets that maximize the revenue.
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Hint.
Find the vertex of the parabola to determine the maximum revenue, then use that to find the price and quantity.
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Answer.
Price: $15, Tickets sold: 150
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Solution.
To find the price and number of tickets that maximize the revenue, we first find the vertex of the parabola represented by the revenue function. The x-coordinate of the vertex can be found using the formula \(x=-\frac{b}{2a}\text{,}\) where \(a=-10\) and \(b=100\text{.}\) Substituting these values, we get \(x=-\frac{100}{2(-10)}=5\text{.}\) This means that the maximum revenue occurs when there are 5 increases in price. The price at this point is \(10+5=15\text{,}\) and the quantity sold is \(200-10(5)=150\text{.}\) Thus, to maximize revenue, they should charge $15 and sell 150 tickets.
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3.

An apple orchard has 30 trees per acre and each tree produces 400 apples per year. The orchard can increase the number of trees by planting more, but for each additional tree planted per acre, the yield per tree decreases by 10 apples. How many trees per acre should the orchard plant to maximize the total apple production per acre?
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(a)
Define the variable for your function.
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Hint.
Pick a variable and state what it represents in the context of the problem.
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Answer.
Let \(x\) represent the number of additional trees planted per acre.
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(b)
Set up the quadratic model.
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Hint.
Write the total apple production function in terms of the variable you defined.
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Answer.
\(P(x)=(30+x)(400-10x)\) or \(P(x)=-10x^2+100x+12000\)
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Solution.
To set up the total apple production function, we first express the number of trees and the yield per tree in terms of \(x\text{.}\) The number of trees can be expressed as \(30+x\text{,}\) since the base number of trees is 30 and it increases by 1 for each additional tree planted. The yield per tree can be expressed as \(400-10x\text{,}\) since each tree produces 400 apples at the base level and produces 10 fewer apples for each additional tree planted. Thus, the total apple production function is: \(P(x)=(30+x)(400-10x)\) or \(P(x)=-10x^2+100x+12000\)
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(c)
State the practical domain.
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Hint.
Consider the context of the problem to determine the practical domain.
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Answer.
\([0,40]\)
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Solution.
Since \(x\) represents the number of additional trees planted per acre, it cannot be negative, so the lower bound of the domain is 0. The upper bound is 40, since if they plant 40 additional trees per acre, each tree will produce 0 apples. Thus, the practical domain is \([0,40]\text{.}\)
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(d)
Find the number of trees per acre that maximize the total apple production.
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Hint.
Find the vertex of the parabola to determine the maximum total apple production, then use that to find the number of trees.
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Answer.
Trees per acre: 5
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Solution.
To find the number of trees per acre that maximize the total apple production, we first find the vertex of the parabola represented by the production function. The x-coordinate of the vertex can be found using the formula \(x=-\frac{b}{2a}\text{,}\) where \(a=-10\) and \(b=100\text{.}\) Substituting these values, we get \(x=-\frac{100}{2(-10)}=5\text{.}\) This means that the maximum total apple production occurs when there are 5 additional trees planted per acre. Thus, to maximize total apple production, they should plant 35 trees per acre.
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4.

A photograph measures 8in by 10in. A border of uniform width is to be added around the photograph. If the area of the border is to be 224 square inches, find the width of the border.
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(a)
Define the variable for your function.
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Hint.
Pick a variable and state what it represents in the context of the problem.
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Answer.
Let \(x\) represent the width of the border in inches.
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(b)
Set up the quadratic model.
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Hint.
Write the area of the border function in terms of the variable you defined. Remember that the area of the border is the area of the larger rectangle (photo + border) minus the area of the photo.
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Answer.
\(A(x)=(8+2x)(10+2x)-80\) or \(A(x)=4x^2+36x\)
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Solution.
To set up the area of the border function, we first express the dimensions of the larger rectangle (photo + border) in terms of \(x\text{.}\) The length of the larger rectangle is \(10+2x\) and the width is \(8+2x\text{.}\) The area of the larger rectangle is \((10+2x)(8+2x)\text{.}\) The area of the photo is 80 square inches. Therefore, the area of the border is \(A(x)=(10+2x)(8+2x)-80\) or \(A(x)=4x^2+36x\text{.}\)
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(c)
Find the width of the border.
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Hint.
Solve the equation \(A(x)=224\) for \(x\text{.}\)
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Answer.
Width of the border: 4 inches
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Solution.
To find the width of the border, we solve the equation \(A(x)=224\) for \(x\text{.}\) Substituting the quadratic model, we get:
\begin{align*} 4x^2+36x &= 224 \\ 4x^2+36x-224 &= 0\\ x^2+9x-56 &= 0\\ (x-4)(x+14) &= 0 \end{align*}
Solving, we find that \(x=4\) or \(x=-14\text{.}\) Since \(x\) represents a width, it must be positive. Therefore, the width of the border is 4 inches.
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