Do all rectangles with the same perimeter, say 36 inches, have the same area? Two different rectangles with perimeter 36 inches are shown. The first rectangle has base 10 inches and height 8 inches, and its area is 80 square inches. The second rectangle has base 12 inches and height 6 inches. Its area is 72 square inches.
The table shows the bases of various rectangles, in inches. Each rectangle has a perimeter of 36 inches. Fill in the height and the area of each rectangle. (To find the height of the rectangle, reason as follows: The base plus the height makes up half of the rectangle’s perimeter.)
What happens to the area of the rectangle when we change its base? On the grid above, plot the points with coordinates (Base, Area). (For this graph we will not use the heights of the rectangles.) The first two points, \((10, 80)\) and \((12, 72)\text{,}\) are shown. Connect your data points with a smooth curve.
Each point on your graph represents a particular rectangle with perimeter 36 inches. The first coordinate of the point gives the base of the rectangle, and the second coordinate gives the area of the rectangle. What is the largest area you found among rectangles with perimeter 36 inches? What is the base for that rectangle? What is its height?
If the rectangle has area 80 square inches, what is its base? Why are there two different answers here? Describe the rectangle corresponding to each answer.
Now we’ll write an algebraic expression for the area of the rectangle in terms of its base. Let \(x\) represent the base of the rectangle. First, express the height of the rectangle in terms of \(x\text{.}\) (Hint: If the perimeter of the rectangle is 36 inches, what is the sum of the base and the height?) Now write an expression for the area of the rectangle in terms of \(x\text{.}\)
Use your formula from part (8) to compute the area of the rectangle when the base is 5 inches. Does your answer agree with the values in your table and the point on your graph?
Use your formula to compute the area of the rectangle when \(x=0\) and when \(x=18\text{.}\) Describe the “rectangles” that correspond to these data points.
The graph of any quadratic equation \(y = ax^2 + bx + c\) is a parabola (if \(a \not= 0\)). In this section we’ll make a systematic study of the parabolas we’ve met earlier in the chapter. Some typical parabolas are shown below.
To begin, notice that these parabolas share certain features.
The graph has either a highest point (if the parabola opens downward, as in figure (a) or a lowest point (if the parabola opens upward, as in figure (b). This high or low point is called the vertex of the graph.
The values of the constants \(a\text{,}\)\(b\text{,}\) and \(c\) determine the location and orientation of the parabola. We’ll consider each of these constants separately.
You can see that the graph of \(y=3x^2\) is narrower than the basic parabola, and the graph of \(y=0.1x^2\) is wider. As \(x\) increases, the graph of \(y=3x^2\) increases faster than the basic parabola, and the graph of \(y=0.1x^2\) increases more slowly. Compare the corresponding \(x\)-values for the three graphs shown in the table.
For each \(x\)-value, the points on the graph of \(y=3x^2\) are higher than the points on the basic parabola, while the points on the graph of \(y=0.1x^2\) are lower. Multiplying by a positive constant greater than 1 stretches the graph vertically, and multiplying by a positive constant less than 1 squashes the graph vertically.
Both equations have the form \(y = ax^2\text{.}\) The graph of \(y = 2x^2\) opens upward because \(a = 2 \gt 0\text{,}\) and the graph of \(y = -\dfrac{1}{2}x^2\) opens downward because \(a = -\dfrac{1}{2}\lt 0\text{.}\)
The graph of \(y=x^2+4\) is shifted upward four units compared to the basic parabola, and the graph of \(y=x^2-4\) is shifted downward four units. Look at the table, which shows the \(y\)-values for the three graphs.
Each point on the graph of \(y=x^2+4\) is four units higher than the corresponding point on the basic parabola, and each point on the graph of \(y=x^2-4\) is four units lower. In particular, the vertex of the graph of \(y=x^2+4\) is the point \((0,4)\text{,}\) and the vertex of the graph of \(y=x^2-4\) is the point \((0,-4)\text{.}\)
The graph of \(y = x^2 - 2\) is shifted downward by two units, compared to the basic parabola. The vertex is the point \((0, -2)\) and the \(x\)-intercepts are the solutions of the equation
\begin{equation*}
0 = x^2 - 2
\end{equation*}
or \(\sqrt{2}\) and \(-\sqrt{2}\text{.}\) The graph is shown below.
The graph of \(y = -x^2 + 4\) opens downward and is shifted \(4\) units up, compared to the basic parabola. Its vertex is the point \((0, 4)\text{.}\) Its \(x\)-intercepts are the solutions of the equation
\begin{equation*}
0 = -x^2 + 4
\end{equation*}
or \(2\) and \(-2\text{.}\) You can verify both graphs with your graphing calculator.
The graph in the standard window is shown below. We see that the axis of symmetry for this parabola is not the \(y\)-axis: the graph is shifted to the left, compared to the basic parabola. We find the \(y\)-intercepts of the graph by setting \(y\) equal to zero: \(~\alert{\text{[TK]}}\)
We can find the vertex of the graph by using the symmetry of the parabola. The \(x\)-coordinate of the vertex lies exactly half-way between the \(x\)-intercepts, so we average their \(x\)-coordinates to find: \(~\alert{\text{[TK]}}\)
Commercial fishermen rely on a steady supply of fish in their area. To avoid overfishing, they adjust their harvest to the size of the population. The formula
\begin{equation*}
R=0.4x-0.001x^2
\end{equation*}
gives the annual rate of growth, in tons per year, of a fish population of biomass \(x\) tons.
Find the vertex of the graph. What does it tell us about the fish population?
where \(v\) is the shuttle’s velocity in ft/sec at touchdown, \(T\) is the pilot’s reaction time before the brakes are applied, and \(a\) is the shuttle’s deceleration.
Suppose that for a particular landing, \(T=0.5\) seconds and \(a=-12\text{ ft/sec}^2\text{.}\) Write the equation for \(d\) in terms of \(v\text{,}\) using these values.
The runway at Edwards Air Force base is 15,000 feet long. Graph your equation again in an appropriate (larger) window, and use it to estimate the answer to the question: What is the maximum velocity the shuttle can have at touchdown and still stop on the runway?
For problems 21–24, graph the set of equations in the standard window on your calculator. Describe how the parameters (constants) \(a\text{,}\)\(b\text{,}\) and \(c\) in each equation transform the graph, compared to the basic parabola.
