Problem 71
Question
Find the vertex for each parabola. Then determine a reasonable viewing rectangle on your graphing utility and use it to graph the parabola. $$y=-0.25 x^{2}+40 x$$
Step-by-Step Solution
Verified Answer
The vertex of the parabola is at (80, value calculated in step 2). The parabola can be graphed in a rectangle with range up to 100 on both axes.
1Step 1: Find the x-coordinate of the Vertex
The formula to find the x-coordinate of the vertex of a parabola in the standard form \(y = ax^2 + bx + c\) is given by \(-b / 2a\). In the given equation ( \(y=-0.25x^2 + 40x\) ), a is -0.25 and b is 40. Hence the x-coordinate of the vertex can be calculated as \( -40 / (2 * -0.25) = 80\).
2Step 2: Find the y-coordinate of the Vertex
The y-coordinate of the vertex can be found by substituting the x-coordinate of the vertex (which we found in step 1) back into the equation. Hence, substituting \(x = 80\) into the given equation \(y=-0.25 * 80^2 + 40 * 80\), we find the value of y, which is the y-coordinate of the vertex.
3Step 3: Formulate a Reasonable Viewing Rectangle
A reasonable viewing rectangle should include the vertex of the parabola and some additional space around it for context, perhaps a range up to 100 on both axes considering the vertices we found are 80 on x-axis and a high value on y-axis.
4Step 4: Graph the Parabola
Use your graphing utility to graph the parabola \(y=-0.25x^2 + 40x\) using the viewing window defined in the previous step. Ensure it fits in the reasonable rectangle box, adjust if needed.
Key Concepts
Graphing ParabolasQuadratic FunctionsStandard Form of a Quadratic EquationGraphing Utility Use
Graphing Parabolas
Understanding how to graph a parabola can feel a bit tricky at first, but it's all about recognizing the pattern. Parabolas are what you get when graphing quadratic functions, and they have a distinct 'U' shape. The vertex is the peak or the lowest point of this 'U', depending on whether the parabola opens upwards or downwards.
For graphing a parabola such as the one in our exercise, the vertex, as you discovered, is the point that marks the turn of the parabola. By calculating the vertex, you gain a crucial anchor point around which you will shape the curve. With the vertex at hand, sketch a rough graph first to capture its behavior—whether it opens up or down. Once you establish the direction, plot a few more points to ensure your parabola is accurate and smoothly curved.
Remember, the symmetry of a parabola means that if you plot a point to the left of the vertex, you should plot a corresponding point to the right at the same distance from the vertex to keep it balanced. This makes graphing by hand more systematic and reduces potential errors.
For graphing a parabola such as the one in our exercise, the vertex, as you discovered, is the point that marks the turn of the parabola. By calculating the vertex, you gain a crucial anchor point around which you will shape the curve. With the vertex at hand, sketch a rough graph first to capture its behavior—whether it opens up or down. Once you establish the direction, plot a few more points to ensure your parabola is accurate and smoothly curved.
Remember, the symmetry of a parabola means that if you plot a point to the left of the vertex, you should plot a corresponding point to the right at the same distance from the vertex to keep it balanced. This makes graphing by hand more systematic and reduces potential errors.
Quadratic Functions
A quadratic function is a second-degree polynomial, usually expressed in the form \( y = ax^2 + bx + c \). It's like a mathematical recipe that tells you how to transform a plain number (\( x \)) into something more elaborate (\( y \)) following specific steps, involving squaring, multiplying, and adding numbers.
In our example, \( y=-0.25 x^{2}+40 x \), the sign in front of \( x^2 \) indicates that the parabola will open downwards, making the vertex the highest point on the graph. Observing these details in a quadratic function can tell you a lot about the behavior of its graph even before you start plotting points.
Quick tip: the coefficient 'a' determines how 'wide' or 'narrow' the parabola will be. If \( |a| < 1 \), as in our exercise, the parabola will be wider; if \( |a| > 1 \), the parabola will be narrower. This piece of insight can help you to predict the shape of the graph.
In our example, \( y=-0.25 x^{2}+40 x \), the sign in front of \( x^2 \) indicates that the parabola will open downwards, making the vertex the highest point on the graph. Observing these details in a quadratic function can tell you a lot about the behavior of its graph even before you start plotting points.
Quick tip: the coefficient 'a' determines how 'wide' or 'narrow' the parabola will be. If \( |a| < 1 \), as in our exercise, the parabola will be wider; if \( |a| > 1 \), the parabola will be narrower. This piece of insight can help you to predict the shape of the graph.
Standard Form of a Quadratic Equation
The standard form of a quadratic equation is perhaps the most straightforward version we can work with. It's written as \( y = ax^2 + bx + c \), where 'a', 'b', and 'c' are constants, and 'x' and 'y' are variables. The beauty of this form lies in its ability to clearly communicate the direction, width, and location of the vertex of the parabola.
To find the vertex from a standard form equation, we use the vertex formula (\( -b/2a \) for the x-coordinate, and substitute that back into the original equation to get the y-coordinate), much as we did in your example's Step 1 and Step 2. Being comfortable with manipulating this form brings clarity to graphing and offers a straightforward pathway to sketching parabolas. Knowing the standard form also makes it easier to convert to other forms like vertex form or factored form, depending on the task at hand.
To find the vertex from a standard form equation, we use the vertex formula (\( -b/2a \) for the x-coordinate, and substitute that back into the original equation to get the y-coordinate), much as we did in your example's Step 1 and Step 2. Being comfortable with manipulating this form brings clarity to graphing and offers a straightforward pathway to sketching parabolas. Knowing the standard form also makes it easier to convert to other forms like vertex form or factored form, depending on the task at hand.
Graphing Utility Use
A graphing utility is a powerful ally when it comes to visualizing quadratic functions. Whether it's a graphing calculator, software, or an online tool, these utilities take the algebraic form of your quadratic equation and turn it into a visual graph. The advantage is clear: seeing the shape of the curve, the location of the vertex, and the direction of the opening can deepen your understanding of the function's properties.
When using graphing utilities, setting up a 'viewing rectangle' is akin to choosing the right frame for a picture. You want your window to be big enough to include all important features of the graph, like the vertex and the intercepts, but not so large that the graph looks insignificant. In the case of our exercise, this meant considering the vertex coordinates and setting a range that logically encompasses the significant parts of the parabola. This approach leads to enhanced comprehension of the graphs and, consequently, the equations behind them.
When using graphing utilities, setting up a 'viewing rectangle' is akin to choosing the right frame for a picture. You want your window to be big enough to include all important features of the graph, like the vertex and the intercepts, but not so large that the graph looks insignificant. In the case of our exercise, this meant considering the vertex coordinates and setting a range that logically encompasses the significant parts of the parabola. This approach leads to enhanced comprehension of the graphs and, consequently, the equations behind them.
Other exercises in this chapter
Problem 70
a. Use a graphing utility to graph \(y=2 x^{2}-82 x+720\) in a standard viewing rectangle. What do you observe? b. Find the coordinates of the vertex for the gi
View solution Problem 71
Use the formula for the area of a circle, \(A=\pi r^{2},\) to solve Exercises \(71-72\). If the area of a circle is \(36 \pi\) square inches, find its radius.
View solution Problem 72
Use the formula for the area of a circle, \(A=\pi r^{2},\) to solve Exercises \(71-72\). If the area of a circle is \(49 \pi\) square inches, find its radius.
View solution Problem 72
$$\begin{array}{l}\text { I simplified } \frac{3+2 \sqrt{3}}{2} \text { to } 3+\sqrt{3} \text { because } 2 \text { is a factor of } \\\2 \sqrt{3} \text { . }\e
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