Quadratic Functions

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

1. Determine the vertex, axis of symmetry, zeros, and \(y\text{-}\) intercept of the parabola shown in .

   i

   The vertex is the turning point of the graph. We can see that the vertex is at \((3,1).\) Because this parabola opens upward, the axis of symmetry is the vertical line that intersects the parabola at the vertex. So the axis of symmetry is \(x=3.\) This parabola does not cross the \(x\text{-}\) axis, so it has no zeros. It crosses the \(y\text{-}\) axis at \((0,7)\) so this is the y-intercept.

2. Write an equation for the quadratic function \(g\) in as a transformation of \(f(x)={x}^{2},\) and then expand the formula, and simplify terms to write the equation in general form.

   i

   We can see the graph of g is the graph of \(f(x)={x}^{2}\) shifted to the left 2 and down 3, giving a formula in the form \(g(x)=a{(x+2)}^{2}-3.\)

   Substituting the coordinates of a point on the curve, such as \((0,-1),\) we can solve for the stretch factor.

   \[\begin{array}{l}-1=a{(0+2)}^{2}-3 \\ 2=4a \\ a=\frac{1}{2}\end{array}\]

   In standard form, the algebraic model for this graph is \(g(x)=\frac{1}{2}{(x+2)}^{2}-3.\)

   To write this in general polynomial form, we can expand the formula and simplify terms.

   \[\begin{array}{l}g(x)=\frac{1}{2}{(x+2)}^{2}-3 \\ =\frac{1}{2}(x+2)(x+2)-3 \\ =\frac{1}{2}({x}^{2}+4x+4)-3 \\ =\frac{1}{2}{x}^{2}+2x+2-3 \\ =\frac{1}{2}{x}^{2}+2x-1\end{array}\]

   Notice that the horizontal and vertical shifts of the basic graph of the quadratic function determine the location of the vertex of the parabola; the vertex is unaffected by stretches and compressions.

3. A coordinate grid has been superimposed over the quadratic path of a basketball in . Assume that the point (–4, 7) is the highest point of the basketball’s trajectory. Find an equation for the path of the ball. Does the shooter make the basket?

   i

   The path passes through the origin and has vertex at \((-4,\ 7),\) so \((h)x=-\frac{7}{16}{(x+4)}^{2}+7.\) To make the shot, \(h(-7.5)\) would need to be about 4 but \(h(-7.5)\approx 1.64;\) he doesn’t make it.

4. Find the vertex of the quadratic function \(f(x)=2{x}^{2}-6x+7.\) Rewrite the quadratic in standard form (vertex form).

   i

   The horizontal coordinate of the vertex will be at

   \[\begin{array}{l}h=-\frac{b}{2a} \\ =-\frac{-6}{2(2)} \\ =\frac{6}{4} \\ =\frac{3}{2}\end{array}\]

   The vertical coordinate of the vertex will be at

   \[\begin{array}{l}k=f(h) \\ =f(\frac{3}{2}) \\ =2{(\frac{3}{2})}^{2}-6(\frac{3}{2})+7 \\ =\frac{5}{2}\end{array}\]

   Rewriting into standard form, the stretch factor will be the same as the \(a\) in the original quadratic.

   \[\begin{array}{l}f(x)=a{x}^{2}+bx+c \\ f(x)=2{x}^{2}-6x+7\end{array}\]

   Using the vertex to determine the shifts,

   \[f(x)=2{(x-\frac{3}{2})}^{2}+\frac{5}{2}\]

5. Given the equation \(g(x)=13+{x}^{2}-6x,\) write the equation in general form and then in standard form.

   i

   \(g(x)={x}^{2}-6x+13\) in general form; \(g(x)={(x-3)}^{2}+4\) in standard form

6. Find the domain and range of \(f(x)=-5{x}^{2}+9x-1.\)

   i

   As with any quadratic function, the domain is all real numbers.

   Because \(a\) is negative, the parabola opens downward and has a maximum value. We need to determine the maximum value. We can begin by finding the \(x\text{-}\) value of the vertex.

   \[\begin{array}{l}h=-\frac{b}{2a} \\ =-\frac{9}{2(-5)} \\ =\frac{9}{10}\end{array}\]

   The maximum value is given by \(f(h).\)

   \[\begin{array}{l}f(\frac{9}{10})=-5{(\frac{9}{10})}^{2}+9(\frac{9}{10})-1 \\ =\frac{61}{20}\end{array}\]

   The range is \(f(x)\le \frac{61}{20},\) or \((-\infty ,\frac{61}{20}].\)

7. Find the domain and range of \(f(x)=2{(x-\frac{4}{7})}^{2}+\frac{8}{11}.\)

   i

   The domain is all real numbers. The range is \(f(x)\ge \frac{8}{11},\) or \([\frac{8}{11},\infty ).\)

8. A backyard farmer wants to enclose a rectangular space for a new garden within her fenced backyard. She has purchased 80 feet of wire fencing to enclose three sides, and she will use a section of the backyard fence as the fourth side.

   1. ⓐ Find a formula for the area enclosed by the fence if the sides of fencing perpendicular to the existing fence have length \(L.\)
   2. ⓑ What dimensions should she make her garden to maximize the enclosed area?
   i

   Let’s use a diagram such as to record the given information. It is also helpful to introduce a temporary variable, \(W,\) to represent the width of the garden and the length of the fence section parallel to the backyard fence.

   1. ⓐ We know we have only 80 feet of fence available, and \(L+W+L=80,\) or more simply, \(2L+W=80.\) This allows us to represent the width, \(W,\) in terms of \(L.\) \[W=80-2L\]

      Now we are ready to write an equation for the area the fence encloses. We know the area of a rectangle is length multiplied by width, so

      \[\begin{array}{l}\ A=LW=L(80-2L) \\ A(L)=80L-2{L}^{2}\end{array}\]

      This formula represents the area of the fence in terms of the variable length \(L.\) The function, written in general form, is

      \[A(L)=-2{L}^{2}+80L.\]
   2. The quadratic has a negative leading coefficient, so the graph will open downward, and the vertex will be the maximum value for the area. In finding the vertex, we must be careful because the equation is not written in standard polynomial form with decreasing powers. This is why we rewrote the function in general form above. Since \(a\) is the coefficient of the squared term, \(a=-2,b=80,\) and \(c=0.\)

   To find the vertex:

   \[\begin{array}{lllll}h=-\frac{80}{2(-2)} & & & & k=A(20) \\ =20 & & \text{and} & & \ =80(20)-2{(20)}^{2} \\ & & & & \ =800\end{array}\]

   The maximum value of the function is an area of 800 square feet, which occurs when \(L=20\) feet. When the shorter sides are 20 feet, there is 40 feet of fencing left for the longer side. To maximize the area, she should enclose the garden so the two shorter sides have length 20 feet and the longer side parallel to the existing fence has length 40 feet.

9. The unit price of an item affects its supply and demand. That is, if the unit price goes up, the demand for the item will usually decrease. For example, a local newspaper currently has 84,000 subscribers at a quarterly charge of $30. Market research has suggested that if the owners raise the price to $32, they would lose 5,000 subscribers. Assuming that subscriptions are linearly related to the price, what price should the newspaper charge for a quarterly subscription to maximize their revenue?

   i

   Revenue is the amount of money a company brings in. In this case, the revenue can be found by multiplying the price per subscription times the number of subscribers, or quantity. We can introduce variables, \(p\) for price per subscription and \(Q\) for quantity, giving us the equation \(\text{Revenue}=pQ.\)

   Because the number of subscribers changes with the price, we need to find a relationship between the variables. We know that currently \(p=30\) and \(Q=84,000.\) We also know that if the price rises to $32, the newspaper would lose 5,000 subscribers, giving a second pair of values, \(p=32\) and \(Q=79,000.\) From this we can find a linear equation relating the two quantities. The slope will be

   \[\begin{array}{l}m=\frac{79,000-84,000}{32-30} \\ =\frac{-5,000}{2} \\ =-2,500\end{array}\]

   This tells us the paper will lose 2,500 subscribers for each dollar they raise the price. We can then solve for the y-intercept.

   \[\begin{array}{ll}\ Q=-2500p+b & \text{Substitute in the point }Q=84,000\text{ and }p=30 \\ 84,000=-2500(30)+b & \text{Solve for }b \\ b=159,000 & \end{array}\]

   This gives us the linear equation \(Q=-2,500p+159,000\) relating cost and subscribers. We now return to our revenue equation.

   \[\begin{array}{l}\text{Revenue}=pQ \\ \text{Revenue}=p(-2,500p+159,000) \\ \text{Revenue}=-2,500{p}^{2}+159,000p\end{array}\]

   We now have a quadratic function for revenue as a function of the subscription charge. To find the price that will maximize revenue for the newspaper, we can find the vertex.

   \[\begin{array}{l}h=-\frac{159,000}{2(-2,500)} \\ =31.8\end{array}\]

   The model tells us that the maximum revenue will occur if the newspaper charges $31.80 for a subscription. To find what the maximum revenue is, we evaluate the revenue function.

   \[\begin{array}{l}\text{maximum revenue}=-2,500{(31.8)}^{2}+159,000(31.8) \\ =2,528,100\end{array}\]

10. Find the y- and x-intercepts of the quadratic \(f(x)=3{x}^{2}+5x-2.\)

    i

    We find the y-intercept by evaluating \(f(0).\)

    \[\begin{array}{l}f(0)=3{(0)}^{2}+5(0)-2 \\ =-2\end{array}\]

    So the y-intercept is at \((0,-2).\)

    For the x-intercepts, we find all solutions of \(f(x)=0.\)

    \[0=3{x}^{2}+5x-2\]

    In this case, the quadratic can be factored easily, providing the simplest method for solution.

    \[0=(3x-1)(x+2)\]\[\begin{array}{lllll}0=3x-1 & & & & 0=x+2 \\ x=\frac{1}{3} & & \text{or} & & x=-2\end{array}\]

    So the x-intercepts are at \((\frac{1}{3},0)\) and \((-2,0).\)

11. Find the \(x\text{-}\) intercepts of the quadratic function \(f(x)=2{x}^{2}+4x-4.\)

    i

    We begin by solving for when the output will be zero.

    \[0=2{x}^{2}+4x-4\]

    Because the quadratic is not easily factorable in this case, we solve for the intercepts by first rewriting the quadratic in standard form.

    \[f(x)=a{(x-h)}^{2}+k\]

    We know that \(a=2.\) Then we solve for \(h\) and \(k.\)

    \[\begin{array}{llll}h=-\frac{b}{2a} & & & k=f(-1) \\ =-\frac{4}{2(2)} & & & \ =2{(-1)}^{2}+4(-1)-4 \\ =-1 & & & \ =-6\end{array}\]

    So now we can rewrite in standard form.

    \[f(x)=2{(x+1)}^{2}-6\]

    We can now solve for when the output will be zero.

    \[\begin{array}{l}0=2{(x+1)}^{2}-6 \\ 6=2{(x+1)}^{2} \\ 3={(x+1)}^{2} \\ x+1=\pm \sqrt{3} \\ x=-1\pm \sqrt{3}\end{array}\]

    The graph has \(x\text{-}\) intercepts at \((-1-\sqrt{3},0)\) and \((-1+\sqrt{3},0).\)

12. In a separate Try It, we found the standard and general form for the function \(g(x)=13+{x}^{2}-6x.\) Now find the y- and \(x\text{-}\) intercepts (if any).

    i

    y-intercept at (0, 13), No \(x\text{-}\) intercepts

13. Solve \({x}^{2}+x+2=0.\)

    i

    Let’s begin by writing the quadratic formula: \(x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}.\)

    When applying the quadratic formula, we identify the coefficients \(a,\ b\text{ and }c.\) For the equation \({x}^{2}+x+2=0,\) we have \(a=1,\ b=1,\ \text{and}\ c=2.\) Substituting these values into the formula we have:
    

    \[\begin{array}{l} \\ x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a} \\ =\frac{-1\pm \sqrt{{1}^{2}-4⋅1⋅(2)}}{2⋅1} \\ =\frac{-1\pm \sqrt{1-8}}{2} \\ =\frac{-1\pm \sqrt{-7}}{2} \\ =\frac{-1\pm i\sqrt{7}}{2}\end{array}\]

    The solutions to the equation are \(\frac{-1+i\sqrt{7}}{2}\) and \(\frac{-1-i\sqrt{7}}{2}\) or \(\frac{-1}{2}+\frac{i\sqrt{7}}{2}\) and \(\frac{-1}{2}-\frac{i\sqrt{7}}{2}.\)

14. A ball is thrown upward from the top of a 40 foot high building at a speed of 80 feet per second. The ball’s height above ground can be modeled by the equation \(H(t)=-16{t}^{2}+80t+40.\)

    1. ⓐ When does the ball reach the maximum height?
    2. ⓑ What is the maximum height of the ball?
    3. ⓒ When does the ball hit the ground?
    i

    1. ⓐ The ball reaches the maximum height at the vertex of the parabola. \[\begin{array}{l}\begin{array}{l} \\ h=-\frac{80}{2(-16)}\end{array} \\ =\frac{80}{32} \\ =\frac{5}{2} \\ =2.5\end{array}\]

       The ball reaches a maximum height after 2.5 seconds.

    2. ⓑ To find the maximum height, find the \(y\text{-}\) coordinate of the vertex of the parabola. \[\begin{array}{l}k=H(-\frac{b}{2a}) \\ =H(2.5) \\ =-16{(2.5)}^{2}+80(2.5)+40 \\ =140\end{array}\]

       The ball reaches a maximum height of 140 feet.

    3. ⓒ To find when the ball hits the ground, we need to determine when the height is zero, \(H(t)=0.\)

       We use the quadratic formula.

       \[\begin{array}{l} \\ t=\frac{-80\pm \sqrt{{80}^{2}-4(-16)(40)}}{2(-16)} \\ =\frac{-80\pm \sqrt{8960}}{-32}\end{array}\]

       Because the square root does not simplify nicely, we can use a calculator to approximate the values of the solutions.

       \[\begin{array}{lll} \\ \\ \begin{array}{lll}t=\frac{-80-\sqrt{8960}}{-32}\approx 5.458 & \text{or} & t=\frac{-80+\sqrt{8960}}{-32}\approx -0.458\end{array}\end{array}\]

       The second answer is outside the reasonable domain of our model, so we conclude the ball will hit the ground after about 5.458 seconds. See

15. A rock is thrown upward from the top of a 112-foot high cliff overlooking the ocean at a speed of 96 feet per second. The rock’s height above ocean can be modeled by the equation \(H(t)=-16{t}^{2}+96t+112.\)

    1. ⓐ When does the rock reach the maximum height?
    2. ⓑ What is the maximum height of the rock?
    3. ⓒWhen does the rock hit the ocean?
    i

    1. ⓐ 3 seconds
    2. ⓑ 256 feet
    3. ⓒ 7 seconds

16. Explain the advantage of writing a quadratic function in standard form.

    i

    When written in that form, the vertex can be easily identified.

17. How can the vertex of a parabola be used in solving real world problems?

18. Explain why the condition of \(a\ne 0\) is imposed in the definition of the quadratic function.

    i

    If \(a=0\) then the function becomes a linear function.

19. What is another name for the standard form of a quadratic function?

20. What two algebraic methods can be used to find the horizontal intercepts of a quadratic function?

    i

    If possible, we can use factoring. Otherwise, we can use the quadratic formula.

21. \(f(x)={x}^{2}-12x+32\)

22. \(g(x)={x}^{2}+2x-3\)

    i

    \(g(x)={(x+1)}^{2}-4,\) Vertex \((-1,-4)\)

23. \(f(x)={x}^{2}-x\)

24. \(f(x)={x}^{2}+5x-2\)

    i

    \(f(x)={(x+\frac{5}{2})}^{2}-\frac{33}{4},\) Vertex \((-\frac{5}{2},-\frac{33}{4})\)

25. \(h(x)=2{x}^{2}+8x-10\)

26. \(k(x)=3{x}^{2}-6x-9\)

    i

    \(f(x)=3{(x-1)}^{2}-12,\) Vertex \((1,-12)\)

27. \(f(x)=2{x}^{2}-6x\)

28. \(f(x)=3{x}^{2}-5x-1\)

    i

    \(f(x)=3{(x-\frac{5}{6})}^{2}-\frac{37}{12},\) Vertex \((\frac{5}{6},-\frac{37}{12})\)

29. \(y(x)=2{x}^{2}+10x+12\)

30. \(f(x)=2{x}^{2}-10x+4\)

    i

    Minimum is \(-\frac{17}{2}\) and occurs at \(\frac{5}{2}.\) Axis of symmetry is \(x=\frac{5}{2}.\)

31. \(f(x)=-{x}^{2}+4x+3\)

32. \(f(x)=4{x}^{2}+x-1\)

    i

    Minimum is \(-\frac{17}{16}\) and occurs at \(-\frac{1}{8}.\) Axis of symmetry is \(x=-\frac{1}{8}.\)

33. \(h(t)=-4{t}^{2}+6t-1\)

34. \(f(x)=\frac{1}{2}{x}^{2}+3x+1\)

    i

    Minimum is \(-\frac{7}{2}\) and occurs at \(-3.\) Axis of symmetry is \(x=-3.\)

35. \(f(x)=-\frac{1}{3}{x}^{2}-2x+3\)

36. \(f(x)={(x-3)}^{2}+2\)

    i

    Domain is \((-\infty ,\infty ).\) Range is \([2,\infty ).\)

37. \(f(x)=-2{(x+3)}^{2}-6\)

38. \(f(x)={x}^{2}+6x+4\)

    i

    Domain is \((-\infty ,\infty ).\) Range is \([-5,\infty ).\)

39. \(f(x)=2{x}^{2}-4x+2\)

40. \(k(x)=3{x}^{2}-6x-9\)

    i

    Domain is \((-\infty ,\infty ).\) Range is \([-12,\infty ).\)

Kuri Gukoresha

Parts of this page are adapted from OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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