Chapter 5: Problem 5
Graph each function. Set the viewing window for \(x\) and \(y\) initially from -5 to 5 then resize if needed. $$y=x^{2}-2 x-1$$
Short Answer
Expert verified
Vertex: (1, -2). X-intercepts: \(1 + \sqrt{2}\) and \(1 - \sqrt{2}\). Graph the parabola with the vertex and x-intercepts, and make sure the parabola opens upwards.
Step by step solution
01
Identify Key Features of the Quadratic Function
The given function, \(y = x^2 - 2x - 1\), is a quadratic function in standard form, \(y = ax^2 + bx + c\), where \(a = 1\), \(b = -2\), and \(c = -1\). Identify the vertex using the formula \(h = -\frac{b}{2a}\) and \(k = c - \frac{b^2}{4a}\).
02
Calculate the Vertex
For \(y = x^2 - 2x - 1\), calculate the vertex: \(h = -\frac{-2}{2 \times 1} = 1\) and \(k = (-1) - \frac{(-2)^2}{4 \times 1} = -1 - 1 = -2\). So the vertex is at (1, -2). This is the lowest point of the parabola since \(a > 0\).
03
Find the x-intercepts if Possible
Set \(y = 0\) to find the x-intercepts. Solve the quadratic equation \(0 = x^2 - 2x - 1\) using factoring, completing the square or the quadratic formula. The discriminant is \(b^2 - 4ac = (-2)^2 - 4(1)(-1) = 8\), since the discriminant is positive, there are two real solutions.
04
Apply the Quadratic Formula
Using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), find the x-intercepts: \(x = \frac{2 \pm \sqrt{8}}{2} = 1 \pm \sqrt{2}\). So the x-intercepts are at \(1 + \sqrt{2}\) and \(1 - \sqrt{2}\).
05
Draw the Graph
Plot the vertex and the x-intercepts on a graph with x and y ranging from -5 to 5. Sketch the parabola opening upwards starting from the vertex, passing through the x-intercepts, and going towards infinity on both sides.
06
Adjust Viewing Window if Necessary
After plotting the initial graph, check if all important features such as the vertex and x-intercepts are visible. If they are not, adjust the viewing window accordingly to ensure all features of the parabola are displayed.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Function
A quadratic function is mathematically represented as \( y = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants with \( a \) not equal to zero. This type of function produces a parabolic graph—the shape of a U or an upside-down U depending on the value of \( a \). If \( a > 0 \), the parabola opens upwards, and if \( a < 0 \), it opens downwards.
When graphing a quadratic function like \( y = x^2 - 2x - 1 \), we first note the coefficients, which give us clues about the function's key characteristics such as its width, direction, and y-intercept. For our function, the \( a \) coefficient is 1, meaning the parabola will be of standard width and open upwards. The \( b \) coefficient is -2, and \( c \) is -1, which affects the position of the parabola on the Cartesian plane. In real-life applications, quadratic functions can model various phenomena, including projectile motion and profit optimization.
When graphing a quadratic function like \( y = x^2 - 2x - 1 \), we first note the coefficients, which give us clues about the function's key characteristics such as its width, direction, and y-intercept. For our function, the \( a \) coefficient is 1, meaning the parabola will be of standard width and open upwards. The \( b \) coefficient is -2, and \( c \) is -1, which affects the position of the parabola on the Cartesian plane. In real-life applications, quadratic functions can model various phenomena, including projectile motion and profit optimization.
Vertex of a Parabola
The vertex of a parabola is a significant feature acting as the peak or the lowest point depending on whether the parabola opens downwards or upwards, respectively. For the given quadratic function \( y = x^2 - 2x - 1 \), we've determined the vertex to be at the point \( (1, -2) \).
Finding the vertex involves using the formulas \( h = -\frac{b}{2a} \) and \( k = c - \frac{b^2}{4a} \), derived from completing the square process of the quadratic equation. In our example, after computation, we found that \( h = 1 \) and \( k = -2 \), signaling that our parabola's vertex lies at \( (1, -2) \). The graph's vertex is crucial for sketching the parabola accurately as it provides a starting point from which the curve extends.
Finding the vertex involves using the formulas \( h = -\frac{b}{2a} \) and \( k = c - \frac{b^2}{4a} \), derived from completing the square process of the quadratic equation. In our example, after computation, we found that \( h = 1 \) and \( k = -2 \), signaling that our parabola's vertex lies at \( (1, -2) \). The graph's vertex is crucial for sketching the parabola accurately as it provides a starting point from which the curve extends.
X-Intercepts
X-intercepts are the points where the graph of the quadratic function crosses the x-axis. These are the values of \( x \) for which \( y \) is zero. To find the x-intercepts of the quadratic function \( y = x^2 - 2x - 1 \), you set the function equal to zero and solve for \( x \).
The discriminant, represented by \( b^2 - 4ac \), tells us the nature of the roots or x-intercepts. A positive discriminant, as we have with the value 8 in our exercise, indicates two real and distinct intercepts. In our step by step example, the x-intercepts were found to be at the points \( 1 + \sqrt{2} \) and \( 1 - \sqrt{2} \), which are crucial in sketching the graph as they define where the parabola meets the x-axis. Without x-intercepts, one might only presume the parabola's path based on its vertex and direction.
The discriminant, represented by \( b^2 - 4ac \), tells us the nature of the roots or x-intercepts. A positive discriminant, as we have with the value 8 in our exercise, indicates two real and distinct intercepts. In our step by step example, the x-intercepts were found to be at the points \( 1 + \sqrt{2} \) and \( 1 - \sqrt{2} \), which are crucial in sketching the graph as they define where the parabola meets the x-axis. Without x-intercepts, one might only presume the parabola's path based on its vertex and direction.
Quadratic Formula
The quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), is a universal solution for the roots of any quadratic equation of the form \( ax^2 + bx + c = 0 \). This powerful formula takes into account the coefficients of the equation and provides the exact values of the x-intercepts when substituted correctly.
In our exercise, applying the quadratic formula allowed us to find the precise x-intercepts \( 1 + \sqrt{2} \) and \( 1 - \sqrt{2} \). Even when the quadratic cannot be factored easily, the quadratic formula is a reliable method for locating where the function will intersect the x-axis. It's essential to correctly substitute the values of \( a \), \( b \), and \( c \) into the formula and carefully perform the arithmetic to ensure accuracy in the resulting intercepts.
In our exercise, applying the quadratic formula allowed us to find the precise x-intercepts \( 1 + \sqrt{2} \) and \( 1 - \sqrt{2} \). Even when the quadratic cannot be factored easily, the quadratic formula is a reliable method for locating where the function will intersect the x-axis. It's essential to correctly substitute the values of \( a \), \( b \), and \( c \) into the formula and carefully perform the arithmetic to ensure accuracy in the resulting intercepts.
