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Section 4.5 Chapter 4 Summary and Review

Subsection 4.5.1 Glossary

  • quadratic formula

  • imaginary unit

  • imaginary number

  • conjugate pair

  • complex number

  • maximum value

  • minimum value

  • vertex form

  • compound inequality

  • interval notation

  • open interval

  • closed interval

Subsection 4.5.2 Key Concepts

  1. The Quadratic Formula.

    The solutions of the equation \(~~ax^2+bx+c=0,~~(a \not= 0)~~\) are

    \begin{equation*} \blert{x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}} \end{equation*}

  2. We have four methods for solving quadratic equations: extraction of roots, factoring, completing the square, and the quadratic formula. The first two methods are faster, but they do not work on all equations. The last two methods work on any quadratic equation.

  3. The imaginary unit, \(i\text{,}\) is defined by \(i=\sqrt{-1}\text{.}\)

  4. The square root of any negative number can be written as the product of a real number (the square root of its absolute value) and \(i\text{.}\)

  5. The sum of a real number and an imaginary number is called a complex number.

  6. The graph of the quadratic equation \(~y=ax^2+bx+c\) may have two, one, or no \(x\)-intercepts, according to the number of distinct real-valued solutions of the equation \(~ax^2+bx+c=0\text{.}\)

  7. The Discriminant.

    The discriminant of a quadratic equation is

    \begin{equation*} D=b^2-4ac \end{equation*}
    1. If \(D \gt 0\text{,}\) there are two unequal real solutions.

    2. If \(D = 0\text{,}\) there is one solution of multiplicity two.

    3. If\(D \lt 0\text{,}\) there are two complex conjugate solutions.

  8. Every quadratic equation has two solutions, which may be the same.

  9. For the graph of \(~y=ax^2+bx+c\text{,}\) the \(x\)-coordinate of the vertex is

    \begin{equation*} x_v=\dfrac{-b}{2a} \end{equation*}
    To find the -coordinate of the vertex, we substitute \(x_v\) into the formula for the parabola.

  10. Quadratic models may arise as the product of two variables.

  11. The maximum or minimum of a quadratic equation occurs at the vertex of its graph.

  12. Vertex Form for a Quadratic Equation.

    A quadratic equation \(~y=ax^2+bx+c,~a \not=0,~\) can be written in the vertex form

    \begin{equation*} \blert{y=a(x-x_v)^2+y_v} \end{equation*}

    where the vertex of the graph is \(~(x_v,y_v)\text{.}\)

  13. We can convert a quadratic equation to vertex form by completing the square.

  14. We need three points to determine a parabola.

  15. We can use the method of elimination to find the equation of a parabola through three points.

  16. If we know the vertex of a parabola, we need only one other point to find its equation.

  17. We can use quadratic regression to fit a parabola to a collection of data points from a quadratic model.

  18. We can use a graphical technique to solve quadratic inequalities.

  19. To solve a quadratic inequality algebraically.

    1. Write the inequality in standard form: One side is zero, and the other has the form \(ax^2+bx+c\text{.}\)

    2. Find the \(x\)-intercepts of the graph of \(~y=ax^2+bx+c\) by setting \(y=0\) and solving for \(x\text{.}\)

    3. Make a rough sketch of the graph, using the sign of \(a\) to determine whether the parabola opens upward or downward.

    4. Decide which intervals on the \(x\)-axis give the correct sign for \(y\text{.}\)

  20. We can write the solution set to a quadratic inequality in interval notation.

  21. Interval Notation.

    1. The closed interval \([a,b]\) is the set \(a \le x \le b\text{.}\)

    2. The open interval \((a,b)\) is the set \(a \lt x \lt b\text{.}\)

    3. Intervals may also be half-open or half-closed.

    4. The infinite interval \([a,\infty)\) is the set \(x \ge a\text{.}\)

    5. The infinite interval \((\infty,a]\) is the set \(x \le a\text{.}\)

  22. To graph a quadratic equation \(~y=ax^2+bx+c\).

    1. Determine whether the parabola opens upward (if \(a \gt 0\)) or downward (if \(a \lt 0\)).

    2. Locate the vertex of the parabola.

      1. The \(x\)-coordinate of the vertex is \(x_v=\dfrac{-b}{2a}\text{.}\)

      2. Find the \(y\)-coordinate of the vertex by substituting \(x_v\) into the equation of the parabola.

    3. Locate the \(x\)-intercepts (if any) by setting \(y=0\) and solving for \(x\text{.}\)

    4. Locate the \(y\)-intercept by evaluating \(y\) for \(x=0\text{.}\)

    5. Locate the point symmetric to the \(y\)-intercept across the axis of symmetry.

Exercises 4.5.3 Chapter 4 Review Problems

Exercise Group.

For Problems 1–4, solve by using the quadratic formula.








\(\dfrac{3 \pm \sqrt{5}}{2}\)




\(-\dfrac{4 \pm \sqrt{8}}{2}\)




\(\dfrac{-2 \pm \sqrt{28}}{4}\)

Exercise Group.

For Problems 5 and 6, solve the formula for the indicated variable.


\(h=6t+3t^2,~~\) for \(t\)


\(\dfrac{6 \pm \sqrt{36-12h}}{6}\)


\(D=\dfrac{n^2-3n}{2},~~\) for \(n\)


\(\dfrac{3 \pm \sqrt{9+8D}}{2}\)

Exercise Group.

For Problems 7–10, use the discriminant to determine the nature of the solutions of the equation.




one repeated real solution




two complex soutions




two complex soutions


\(\dfrac{x^2}{4} = x+\dfrac{5}{4}\)


two distinct real soutions


The height, \(h\text{,}\) of an object \(t\) seconds after being thrown from ground level is given by

\begin{equation*} h=v_0 t-\dfrac{1}{2}gt^2 \end{equation*}

where \(v_0\) is its starting velocity and \(g\) is a constant that depends on gravity. On the moon, the value of \(g\) is approximately 5.6. Suppose you hit a golf ball on the moon with an upwards velocity 100 feet per second.

  1. Write an equation for the height of the golf ball \(t\) seconds after you hit it.

  2. Graph your equation in the window

    \begin{align*} \text{Xmin} \amp = 0 \amp\amp \text{Xmax} = 47\\ \text{Ymin} \amp = 0 \amp\amp \text{Ymax} = 1000 \end{align*}

  3. Use the TRACE key to estimate the maximum height the golf ball reaches.

  4. Use your equation to calculate when the golf ball will reach a height of 880 feet.

  1. \(\displaystyle h=100t-2.8t^2\)

  2. (graph)

  3. 893 ft

  4. \(15 \dfrac{5}{7}\) sec and 20 sec


An acrobat is catapulted into the air from a springboard at ground level. His height \(h\) in meters is given by the formula

\begin{equation*} h=-4.9t^2+14.7t \end{equation*}

where \(t\) is the time in seconds from launch. Use your calculator to graph the acrobat's height versus time. Set the WINDOW values on your calculator to

\begin{align*} \text{Xmin} \amp = 0 \amp\amp \text{Xmax} = 4.7\\ \text{Ymin} \amp = 0 \amp\amp \text{Ymax} = 12 \end{align*}
  1. Use the TRACE key to find the coordinates of the highest point on the graph. When does the acrobat reach his maximum height, and what is that height?

  2. Use the formula to find the height of the acrobat after 2.4 seconds.

  3. Use the TRACE key to verify your answer to part (b). Find another time when the acrobat is at the same height.

  4. Use the formula to find two times when the acrobat is at a height of 6.125 meters. Verify your answers on the graph.

  5. What are the coordinates of the horizontal intercepts of your graph? What do these points have to do with the acrobat?

  1. 1.5 sec, 11.025 m

  2. 7.056 m

  3. 0.6 sec

  4. 0.5 sec, 2.5 sec

  5. \((0,0),~(3,0).\) She leaves the springboard at \(t=0\) seconds and returns to the springboard at \(t=3\) seconds.

Exercise Group.

For Problems 13-16,

  1. Find the coordinates of the vertex and the intercepts.

  2. Sketch the graph.



  1. \(\left(\dfrac{1}{2}, \dfrac{-49}{4}\right)\text{,}\) \((-3,0)\text{,}\) \((4,0)\text{,}\) \((0,-12)\)

  2. parabola


  1. \(\left(\dfrac{1}{4}, \dfrac{-31}{8}\right)\text{,}\) no \(x\)-intercepts,\(~(0,-4)\)

  2. parabola


  1. \(\displaystyle (1,5),~(-1.24,0),~(3.24,0),~(0,4)\)

  2. parabola


  1. \(\left(\dfrac{3}{2}, \dfrac{7}{4}\right),~\) no \(x\)-intercepts,\(~(0,4)\)

  2. parabola


Find the equation for a parabola whose vertex is \((15,-6)\) and that passes through the point \((3,22.8)\text{.}\)




  1. Find the vertex of the graph of \(y=-2(x-1)^2+5\)

  2. Write the equation of the parabola in standard form.

  1. \(\displaystyle (1,5)\)

  2. \(\displaystyle y=2x^2+4x+7\)


The total profit Kiyoshi makes from producing and selling \(x\) floral arrangements is

\begin{equation*} P=-0.4x^2+36x \end{equation*}
  1. How many floral arrangements should Kiyoshi produce and sell to maximize his profit?

  2. What is his maximum profit?

  3. Verify your answers on a graph.

  1. 45

  2. $810


The Metro Rail service sells \(1200-80x\) fares each day when it charges \(x\) dollars per fare.

  1. Write an equation for the revenue in terms of the price of a fare.

  2. What fare will return the maximum revenue?

  3. What is the maximum revenue?

  4. Verify your answers on a graph.

  1. \(\displaystyle R=1200x-80x^2\)

  2. $7.50

  3. $4500


A beekeeper has beehives distributed over 60 square miles of pastureland. When she places four hives per square mile, each hive produces about 32 pints of honey per year. For each additional hive per square mile, honey production drops by 4 pints per hive.

  1. Write an equation for the total production of honey, in pints, in terms of the number of additional hives per square mile.

  2. How many additional hives per square mile should the beekeeper install in order to maximize honey production?

  1. \(\displaystyle y=60(4+2x)(32-4x)\)

  2. 2


A small company manufactures radios. When it charges $20 for a radio, it sells 500 radios per month. For each dollar the price is increased, 10 fewer radios are sold per month.

  1. Write an equation for the monthly revenue in terms of the price increase over $20.

  2. What should the company charge for a radio in order to maximize its monthly revenue?

  1. \(\displaystyle R=(20+x)(500-10x)\)

  2. $35


Find values of \(a~, b,\) and \(c\) so that the graph of the parabola \(~y=ax^2+bx+c\) contains the points \((-1,-4),~ (0,-6)~\) and \((4,6)\text{.}\)




Find a parabola that fits the following data points.

\(x\) \(-8\) \(-4\) \(2\) \(4\)
\(y\) \(10\) \(18\) \(0\) \(-14\)



The height of a cannonball was observed at \(0.2\)-second intervals after the cannon was fired, and the data recorded in the table below.

Time (seconds) \(0.2\) \(0.4\) \(0.6\) \(0.8\) \(1.0\) \(1.2\) \(1.4\) \(1.6\) \(1.8\) \(2.0\)
Height(metes) \(10.2\) \(19.2\) \(27.8\) \(35.9\) \(43.7\) \(51.1\) \(58.1\) \(64.7\) \(71.0\) \(76.8\)
  1. Find the equation of the least-squares regression line for height in terms of time.

  2. Use the linear regression equation to predict the height of the cannonball at 3 seconds and at 4 seconds after it was fired.

  3. Make a scatterplot of the data and draw the regression line on the same axes.

  4. Find the quadratic regression equation for height in terms of time.

  5. Use quadratic regression equation to predict the height of the cannonball at 3 seconds and at 4 seconds after it was fired.

  6. Make a scatterplot of the data and draw the regression curve on the same axes.

  7. Which model is more appropriate for the height of the cannonball, linear or quadratic? Why?

  1. \(\displaystyle h=36.98t+5.17\)

  2. 116.1 m, 153.1 m

  3. (graph)

  4. \(\displaystyle h=-4.858t^2+47.67t+0.89\)

  5. 100.2 m, 113.9 m

  6. (graph)

  7. quadratic


Max took a sequence of photographs of an explosion spaced at equal time intervals. From the photographs he was able to estimate the height and vertical velocity of some debris from the explosion, as shown in the table. (Negative velocities indicate that the debris is falling back to earth.)

Velocity (meters/second) \(67\) \(47\) \(27\) \(8\) \(-12\) \(-31\)
Height (meters) \(8\) \(122\) \(196\) \(232\) \(228\) \(185\)
  1. Enter the data into your calculator and create a scatterplot. Fit a quadratic regression equation to the data, and graph the equation on the scatterplot.

  2. Use your regression equation to find the vertex of the parabola. What do the coordinates represent, in terms of the problem? What should the velocity of the debris be at its maximum height?

  1. \(\displaystyle y=-0.05x^2-0.003x+234.2\)

  2. \((-0.03, 234.2)~\) The velocity of the debris at its maximum height of 234.2 feet. The velocity there is actually zero.

Exercise Group.

For Problems 27–32, solve the inequality algebraically, and give your answers in interval notation. Verify your solutions by graphing.


\((x-3)(x+2) \gt 0\)


\((-\infty,-2) \cup (3,\infty)\)


\(y^2-y-12 \le 0\)




\(2y^2-y \le 3\)




\(3z^2-5z \gt 2\)


\((-\infty,-\dfrac{1}{3}) \cup (2,\infty)\)


\(s^2 \le 4\)




\(4t^2 \gt 12\)


\((-\infty,-\sqrt{3}) \cup (\sqrt{3},\infty)\)


The Sub Station sells \(~220-\dfrac{1}{4}p~\) submarine sandwiches at lunchtime if it sells them at \(p\) cents each.

  1. Write a formula for the Sub Station's daily revenue in terms of \(p\text{.}\)

  2. What range of prices can the Sub Station charge if it wants to keep its daily revenue from subs over $480?

  1. \(\displaystyle R=p(220-\dfrac{1}{4}p)\)

  2. \(\displaystyle 400 \lt p \lt 480\)


When it charges \(p\) dollars for an electric screwdriver, Handy Hardware will sell \(~30-\dfrac{1}{2}p~\) screwdrivers per month.

  1. Write a formula in terms of \(p\) for Handy Hardware's monthly revenue from screwdrivers.

  2. How much should Handy charge per screwdriver if it wants the monthly revenue from the screwdrivers to be over $400?

  1. \(\displaystyle R=p(30-\dfrac{1}{2}p)\)

  2. \(\displaystyle 20 \lt p \lt 40\)

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