The profit on your school fundraiser is represented by the quadratic expression \begin{align*}-3p^2 + 200p - 3000\end{align*} , where p is your price point. What price point will result in the maximum profit and what is that maximum profit? Hint: Find the vertex of the parabola.
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James Sousa: Ex 4: Graph a Quadratic Function in General Form by Finding Key Components
Guidance
Now that we have found the solutions of a quadratic equation we will graph the function. First, we need to introduce @$\begin{align*}y\end{align*}@$ or @$\begin{align*}f(x)\end{align*}@$ . A quadratic function is written @$\begin{align*}y=ax^2+bx+c\end{align*}@$ or @$\begin{align*}f(x)=ax^2+bx+c\end{align*}@$ (see Finding the Domain and Range of Functions concept). All quadratic equations are also functions.
Recall that the solutions of a quadratic equation are found when the equation is set equal to zero. This is also the same as when @$\begin{align*}y = 0\end{align*}@$ . Therefore, the solutions of a quadratic equation are also the @$\begin{align*}{\color{red}x-\mathbf{intercepts}}\end{align*}@$ of that function, when graphed.
The graph of a quadratic equation is called a @$\begin{align*}{\color{blue}\mathbf{parabola}}\end{align*}@$ and looks like the figure to the left. A parabola always has “U” shape and depending on certain values, can be wider or narrower. The lowest part of a parabola, or minimum, is called the @$\begin{align*}{\color{green}\mathbf{vertex}}\end{align*}@$ . Parabolas can also be flipped upside-down, and in this case, the vertex would be the maximum value. Notice that this parabola is symmetrical about vertical line that passes through the vertex. This line is called the axis of symmetry. Lastly, where the parabola passes through the @$\begin{align*}y-\end{align*}@$ axis (when @$\begin{align*}x = 0\end{align*}@$ ), is the @$\begin{align*}{\color{magenta}y-\mathbf{intercept}}\end{align*}@$ .
If you are given (or can find) the @$\begin{align*}x-\end{align*}@$ intercepts and the vertex, you can always graph a parabola.
Investigation: Finding the Vertex of a Parabola
1. The equation of the parabola above is @$\begin{align*}y=x^2-2x-3\end{align*}@$ . Find @$\begin{align*}a, b,\end{align*}@$ and @$\begin{align*}c\end{align*}@$ . @$\begin{align*}a=1,b=-2,c=-3\end{align*}@$
2. What are the coordinates of the vertex? (1, -4)
3. Create an expression using @$\begin{align*}a\end{align*}@$ and @$\begin{align*}b\end{align*}@$ (from Step 1) that would be equal to the @$\begin{align*}x-\end{align*}@$ coordinate of the vertex. @$\begin{align*}1=\frac{-b}{2a}\end{align*}@$
4. Plug in @$\begin{align*}x = 1\end{align*}@$ to the equation of the parabola. What do you get for @$\begin{align*}y\end{align*}@$ ? @$\begin{align*}y = -4\end{align*}@$
From this investigation, we have introduced how to find the vertex of a parabola. The @$\begin{align*}x-\end{align*}@$ coordinate of the vertex is @$\begin{align*}x=\frac{-b}{2a}\end{align*}@$ . To find @$\begin{align*}y\end{align*}@$ , plug in this value to the equation, also written @$\begin{align*}f\left(\frac{-b}{2a}\right)\end{align*}@$ . @$\begin{align*}x=\frac{-b}{2a}\end{align*}@$ is also the equation of the axis of symmetry.
Example A
Find the vertex, axis of symmetry, @$\begin{align*}x-\end{align*}@$ intercepts, and @$\begin{align*}y-\end{align*}@$ intercept of @$\begin{align*}y=-\frac{1}{2}x^2-2x+6\end{align*}@$ .
Solution: First, let’s find the @$\begin{align*}x-\end{align*}@$ intercepts. This equation is factorable and @$\begin{align*}ac = -3\end{align*}@$ . The factors of -3 that add up to -2 are -3 and 1. Expand the @$\begin{align*}x-\end{align*}@$ term and factor.
@$$\begin{align*}-\frac{1}{2}x^2-2x+6 &=0\\ -\frac{1}{2}x^2-3x+x+6 &=0\\ -x\left(\frac{1}{2}x+3\right)+2\left(\frac{1}{2}x+3\right) &=0\\ \left(\frac{1}{2}x+3\right)(-x+2) &=0\end{align*}@$$
Solving for @$\begin{align*}x\end{align*}@$ , the intercepts are (-6, 0) and (2, 0).
To find the vertex, use @$\begin{align*}x=\frac{-b}{2a}\end{align*}@$ .
@$\begin{align*}x=\frac{-(-2)}{2 \cdot -\frac{1}{2}}=\frac{2}{-1}=-2\end{align*}@$ Plug this into the equation: @$\begin{align*}y=-\frac{1}{2}(-2)^2-2(-2)+6=-2+4+6=8\end{align*}@$ .
Therefore, the vertex is (-2, 8) and the axis of symmetry is @$\begin{align*}x = -2\end{align*}@$ .
To find the @$\begin{align*}y-\end{align*}@$ intercept, @$\begin{align*}x = 0\end{align*}@$ . @$\begin{align*}y=-\frac{1}{2}(0)^2-2(0)+6=6\end{align*}@$ . Therefore, the @$\begin{align*}y-\end{align*}@$ intercept is (0, 6).
Example B
Sketch a graph of the parabola from Example A.
Solution: Plot the vertex and two @$\begin{align*}x-\end{align*}@$ intercepts (red points). Plot the @$\begin{align*}y-\end{align*}@$ intercept. Because all parabolas are symmetric, the corresponding point on the other side would be (-4, 6). Connect the five points to form the parabola.
For this parabola, the vertex is the maximum value. If you look at the equation, @$\begin{align*}y=-\frac{1}{2}x^2-2x+6\end{align*}@$ , we see that the @$\begin{align*}a\end{align*}@$ value is negative. When @$\begin{align*}a\end{align*}@$ is negative, the sides of the parabola, will point down.
Example C
Find the vertex and @$\begin{align*}x-\end{align*}@$ intercepts of @$\begin{align*}y=2x^2-5x-25\end{align*}@$ . Then, sketch a graph.
Solution: First, this is a factorable function. @$\begin{align*}ac = -50\end{align*}@$ . The factors of -50 that add up to -5 are -10 and 5.
@$$\begin{align*}2x^2-5x-25 &=0\\ 2x^2-10x+5x-25 &=0\\ 2x(x-5)+5(x-5) &=0\\ (2x+5)(x-5) &=0\end{align*}@$$
Setting each factor equal to zero, we get @$\begin{align*}x = 5\end{align*}@$ and @$\begin{align*}-\frac{5}{2}\end{align*}@$ .
From this, we get that the @$\begin{align*}x-\end{align*}@$ intercepts are (5, 0) and @$\begin{align*}\left(-\frac{5}{2},0\right)\end{align*}@$ . To find the vertex, use @$\begin{align*}x=\frac{-b}{2a}\end{align*}@$ .
@$\begin{align*}x = \frac{5}{2 \cdot 2}=\frac{5}{4}\end{align*}@$ Now, find @$\begin{align*}y\end{align*}@$ . @$\begin{align*}y=2\left(\frac{5}{4}\right)^2-5\left(\frac{5}{4}\right)-25=\frac{25}{8}-\frac{25}{4}-25=-\frac{225}{8}=-28\frac{1}{8}\end{align*}@$
The vertex is @$\begin{align*}\left(\frac{5}{4}, -28\frac{1}{8}\right)\end{align*}@$ . To graph this, we will need to estimate the vertex and draw an appropriate scale for the grid. As a decimal, the vertex is (1.25, -28.125).
Intro Problem Revisit The maximum profit occurs that the maximum point of the parabola, so find the vertex of @$\begin{align*}-3p^2 + 200p - 3000\end{align*}@$ .
To find the vertex, use @$\begin{align*}x=\frac{-b}{2a}\end{align*}@$ .
@$\begin{align*}x=\frac{-(200)}{2 \cdot -3}=\frac{-200}{-6}=33.33\end{align*}@$ Plug this into the equation: @$\begin{align*}y=(-3)(33.33)^2 +(200)(33.33) -3000=-3333+6667-3000=334\end{align*}@$ .
Therefore, the vertex is (33.33, 334) and the maximum profit occurs at a price point of $33.33. At that price point, the profit would be $334.
Guided Practice
1. Find the @$\begin{align*}x-\end{align*}@$ intercepts, @$\begin{align*}y-\end{align*}@$ intercept, vertex, and axis of symmetry of @$\begin{align*}y=-x^2+7x-12\end{align*}@$ .
2. Sketch a graph of the parabola from #1.
3. Find the vertex of @$\begin{align*}y=-4x^2+16x-17\end{align*}@$ . Does the parabola open up or down?
Answers
1. This is a factorable quadratic equation.
@$$\begin{align*}-(x^2-7x+12) &=0\\ -(x^2-3x-4x+12) &=0\\ -[x(x-3)-4(x-3)] &=0\\ -(x-3)(x-4) &=0\end{align*}@$$
The @$\begin{align*}x-\end{align*}@$ intercepts are (3, 0) and (4, 0).
@$$\begin{align*}y &=-0^2+7(0)-12\\ y &=-12\end{align*}@$$
The @$\begin{align*}y-\end{align*}@$ intercept is (0, -12).
The @$\begin{align*}x-\end{align*}@$ coordinate of the vertex is @$\begin{align*}x=\frac{-7}{2(-1)}=\frac{7}{2}\end{align*}@$ . The @$\begin{align*}y-\end{align*}@$ coordinate is @$\begin{align*}y=-\left(\frac{7}{2}\right)^2+7\left(\frac{7}{2}\right)-12=\frac{1}{4}\end{align*}@$ .
Therefore, the vertex is @$\begin{align*}\left(\frac{7}{2}, \frac{1}{4}\right)\end{align*}@$ and the parabola opens down because @$\begin{align*}a<0\end{align*}@$ . The axis of symmetry is @$\begin{align*}x=\frac{7}{2}\end{align*}@$ .
2. Plot all the points you found in #1. Then, connect the points to create the parabola.
3. First, the parabola opens down because @$\begin{align*}a\end{align*}@$ is negative. The @$\begin{align*}x-\end{align*}@$ coordinate of the vertex is @$\begin{align*}x=\frac{-16}{2(-4)}=\frac{-16}{-8}=2\end{align*}@$ . The @$\begin{align*}y-\end{align*}@$ coordinate is @$\begin{align*}y=-4(2)^2+16(2)-17=-16+32-17=-1\end{align*}@$ . This makes the vertex (2, -1).
Even though the problem does not ask, we can infer that this parabola does not cross the @$\begin{align*}x-\end{align*}@$ axis because it points down and the vertex is below the @$\begin{align*}x-\end{align*}@$ axis. This means that the solutions would be imaginary.
Explore More
Find the vertex of each parabola and determine if it is a maximum or minimum.
- @$\begin{align*}y=x^2-12x+11\end{align*}@$
- @$\begin{align*}y=x^2+10x-18\end{align*}@$
- @$\begin{align*}y=-3x^2+4x+17\end{align*}@$
- @$\begin{align*}y=2x^2-9x-11\end{align*}@$
- @$\begin{align*}y=-x^2+6x-9\end{align*}@$
- @$\begin{align*}y=-\frac{1}{4}x^2+8x-33\end{align*}@$
Find the vertex, @$\begin{align*}x-\end{align*}@$ intercepts, @$\begin{align*}y-\end{align*}@$ intercept, and axis of symmetry of each factorable quadratic equation below. Then, sketch a graph of each one.
- @$\begin{align*}y=x^2-12x+11\end{align*}@$
- @$\begin{align*}y=-2x^2-5x+12\end{align*}@$
- @$\begin{align*}y=\frac{1}{3}x^2+4x-15\end{align*}@$
- @$\begin{align*}y=3x^2+26x-9\end{align*}@$
- @$\begin{align*}y=-x^2+10x-25\end{align*}@$
- @$\begin{align*}y=-\frac{1}{2}x^2+5x+28\end{align*}@$
- If a function is not factorable, how would you find the @$\begin{align*}x-\end{align*}@$ intercepts?
Find the vertex and @$\begin{align*}x-\end{align*}@$ intercepts of the following quadratic equations. Then, sketch the graph. These equations are not factorable.
- @$\begin{align*}y=-x^2+8x-9\end{align*}@$
- @$\begin{align*}y=2x^2-x-8\end{align*}@$
Complete the table of values for the quadratic equations below. Then, plot the points and graph.
- @$\begin{align*}y=x^2-2x+2\end{align*}@$ , x = 5, 3, 1, -1, -3, and -5
- @$\begin{align*}y=x^2+4x+13\end{align*}@$ , x = 4, 0, -2, -4, and -8
- Writing What do you notice about the two parabolas from 16 and 17? What type of solutions do these functions have? Solve #16.
- Writing How many different ways can a parabola intersect the @$\begin{align*}x-\end{align*}@$ axis? Draw parabolas on an @$\begin{align*}x-y\end{align*}@$ plane to represent the different solution possibilities.
- Challenge If the @$\begin{align*}x-\end{align*}@$ coordinate of the vertex is @$\begin{align*}\frac{-b}{2a}\end{align*}@$ for @$\begin{align*}y=ax^2+bx+c\end{align*}@$ , find the @$\begin{align*}y-\end{align*}@$ coordinate in terms of @$\begin{align*}a, b,\end{align*}@$ and @$\begin{align*}c\end{align*}@$ .