Chapter 13: Problem 50
Sketch the graph of each equation. If the graph is a parabola, find irs vertex. If the graph is a circle, find its center and radius. $$y=x^{2}-2 x-15$$
Short Answer
Expert verified
The graph is a parabola with vertex at (1, -16).
Step by step solution
01
Identify the Equation Type
The exercise provided is the equation \(y = x^2 - 2x - 15\). This equation is of the form \(y = ax^2 + bx + c\), which is a quadratic equation. This indicates that the graph is a parabola, as quadratic equations graphically represent parabolas.
02
Find the Vertex
For a quadratic equation in the form \(y = ax^2 + bx + c\), the vertex can be found using the formula \(x = -\frac{b}{2a}\). Here, \(a = 1\), \(b = -2\), and \(c = -15\). So, the x-coordinate of the vertex is \(-\frac{-2}{2 \times 1} = 1\). Substitute \(x = 1\) back into the equation to find the y-coordinate: \(y = (1)^2 - 2(1) - 15 = -16\). Thus, the vertex of the parabola is \((1, -16)\).
03
Plot the Vertex
On a coordinate plane, plot the vertex of the parabola at point \((1, -16)\). This point is a crucial starting point for sketching the graph of the parabola.
04
Determine the Parabola's Orientation
Since the coefficient of \(x^2\) (\(a = 1\)) is positive, the parabola opens upwards. This means that the parabola will have a minimum point at its vertex.
05
Sketch the Parabola
Using the vertex \((1, -16)\), draw the U-shaped parabola that opens upward. The vertex is the lowest point, and the parabola is symmetric with respect to the vertical line that passes through the vertex at \(x = 1\). Additional points can be plotted by choosing values for \(x\) on either side of the vertex and calculating corresponding \(y\) values.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Parabola
In mathematics, a parabola is a symmetrical, U-shaped curve. When we're dealing with quadratic equations of the form \(y = ax^2 + bx + c\), we're essentially talking about parabolas. The direction in which a parabola opens is determined by the sign of the coefficient \(a\). If \(a > 0\), the parabola opens upward, forming a U shape. Conversely, if \(a < 0\), it opens downward, resembling an upside-down U.
Parabolas have certain distinctive characteristics:
Parabolas have certain distinctive characteristics:
- They are mirror symmetric around a perpendicular line that goes through the vertex.
- The vertex is either the lowest or highest point, depending on the parabola's orientation.
Vertex
The vertex of a parabola is an essential feature as it represents the peak or the trough of the curve. For the quadratic equation given, \(y = x^2 - 2x - 15\), finding the vertex involves a bit of calculation but is quite straightforward.
The x-coordinate of the vertex is found using the formula \(x = -\frac{b}{2a}\). Here, \(b = -2\) and \(a = 1\), so the x-coordinate is \(1\). Next, to find the y-coordinate, we substitute this x-value back into the equation, resulting in ; y = (1)^2 - 2(1) - 15 = -16. Thus, the vertex of this parabola is located at \( (1, -16) \).
The vertex represents the lowest point on this upward-opening parabola, often referred to as a minimum point.
The x-coordinate of the vertex is found using the formula \(x = -\frac{b}{2a}\). Here, \(b = -2\) and \(a = 1\), so the x-coordinate is \(1\). Next, to find the y-coordinate, we substitute this x-value back into the equation, resulting in ; y = (1)^2 - 2(1) - 15 = -16. Thus, the vertex of this parabola is located at \( (1, -16) \).
The vertex represents the lowest point on this upward-opening parabola, often referred to as a minimum point.
Graphing
Graphing a quadratic equation consists of plotting its parabola on a coordinate plane. The first step is to mark the vertex, which serves as a foundation for the graph. For the equation \(y = x^2 - 2x - 15\), we plot the vertex at \( (1, -16) \).
Next, identify the direction of the parabola's opening, which we've determined opens upwards since \(a=1\) is positive. This step guarantees the correct orientation of the U shape on the graph.
The parabola is symmetric about the vertical line \(x=1\), ensuring both sides of our graph will mirror each other.
To complete the graph, it's beneficial to select additional x-values on either side of the vertex, compute their corresponding y-values, and plot these points. This approach assists in refining the shape of the parabola.
Next, identify the direction of the parabola's opening, which we've determined opens upwards since \(a=1\) is positive. This step guarantees the correct orientation of the U shape on the graph.
The parabola is symmetric about the vertical line \(x=1\), ensuring both sides of our graph will mirror each other.
To complete the graph, it's beneficial to select additional x-values on either side of the vertex, compute their corresponding y-values, and plot these points. This approach assists in refining the shape of the parabola.
Quadratic Formula
The quadratic formula is a powerful tool for solving quadratic equations. Although not directly used in graphing, it provides profound insight into the behavior of quadratic equations, including where their parabolas intersect the x-axis. The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Using this formula, you can determine the roots or zeros of a quadratic equation, which are the x-values where the parabola intersects the x-axis. The portion inside the square root, \(b^2 - 4ac\), is known as the discriminant.
The discriminant reveals even more:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Using this formula, you can determine the roots or zeros of a quadratic equation, which are the x-values where the parabola intersects the x-axis. The portion inside the square root, \(b^2 - 4ac\), is known as the discriminant.
The discriminant reveals even more:
- If \(b^2 - 4ac > 0\), there are two distinct real roots.
- If \(b^2 - 4ac = 0\), there is one real root or a double root.
- If \(b^2 - 4ac < 0\), the equation has no real roots, implying the parabola does not cross the x-axis.