Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 8: Problem 43
Find the vertex and intercepts for each quadratic function. Sketch the graph, and state the domain and range. $$f(x)=x^{2}-x-2$$
Short Answer
Expert verified
Vertex: \(\left( \frac{1}{2}, -\frac{9}{4} \right)\), x-intercepts: \(2, 0\) and \(-1, 0\), y-intercept: \(0, -2\), Domain: \((-\infty, \,\infty)\), Range: \([-\frac{9}{4}, \,\infty)\).
Step by step solution
01
Rewrite in Standard Form
The quadratic function is already given in the form \[ f(x) = x^2 - x - 2.\]
02
Find the Vertex
To find the vertex, use the formula \[ x = -\frac{b}{2a}.\] For the quadratic function \[ f(x)= x^2 - x - 2,\] we have \[ a=1, b=-1.\] So, \[ x = -\frac{-1}{2(1)} = \frac{1}{2}.\] Now plug this value into the original equation to find \[ y: \] \[ y = \left(\frac{1}{2}\right)^2 - \frac{1}{2} - 2 = \frac{1}{4} - \frac{1}{2} - 2 = \frac{1}{4} - \frac{2}{4} - \frac{8}{4} = -\frac{9}{4}.\] So, the vertex is \left(\frac{1}{2}, -\frac{9}{4}\right).
03
Find the x-intercepts
To find the x-intercepts, set \[ f(x) = 0\] and solve for \[ x:\]\[ x^2 - x - 2 = 0.\] Factoring the quadratic equation, we get \[ (x - 2)(x + 1) = 0.\] Thus, \[ x = 2\] or \[ x = -1.\] So, the x-intercepts are \[(2,0)\] and \[(-1,0).\]
04
Find the y-intercept
To find the y-intercept, set \[ x = 0\] in the equation and solve for \[ f(0):\]\[ f(0) = 0^2 - 0 - 2 = -2.\] So, the y-intercept is \[(0, -2).\]
05
Sketch the Graph
Plot the vertex \[\left(\frac{1}{2}, -\frac{9}{4}\right),\] the x-intercepts \[(2,0)\] and \[(-1,0),\] and the y-intercept \[(0, -2).\] Draw a parabola that opens upwards passing through these points.
06
State the Domain
The quadratic function is defined for all real numbers, so the domain is \[ (-\infty, \infty). \]
07
State the Range
Since the parabola opens upwards and the vertex is \(\left( \frac{1}{2}, -\frac{9}{4} \right)\),the range is \[ \left[-\frac{9}{4}, \infty\right).\]
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 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
quadratic function
A quadratic function is a type of polynomial function where the highest degree of the variable is 2. The general form of a quadratic function is written as \[ f(x) = ax^2 + bx + c. \] In this equation:
- \( a \) is the coefficient of \( x^2 \),
- \( b \) is the coefficient of \( x \)
- \( c \) is the constant term.
- If \( a > 0 \), the parabola opens upwards.
- If \( a < 0 \), the parabola opens downwards.
vertex
The vertex of a quadratic function is a key point on its graph. It represents the minimum or maximum point of the parabola, depending on its direction. To find the vertex of a quadratic function in the form \[ f(x) = ax^2 + bx + c, \] use the formula: \[ x = -\frac{b}{2a}. \] Plug this value of \( x \) back into the function to find the corresponding \( y \)-value. Together, these values \( (x, y) \) are the vertex coordinates. For example, in the function \[ f(x) = x^2 - x - 2, \]
- \( a = 1, \)
- \( b = -1. \)
- \(\therefore \,x = -\frac{-1}{2(1)} = \frac{1}{2}.\)
x-intercepts
The x-intercepts of a quadratic function are the points where the graph crosses the x-axis. At these points, the function's value is zero (\( f(x) = 0 \)). To find the x-intercepts, solve the equation \[ ax^2 + bx + c = 0. \] You can factorize the quadratic equation or use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \] For example, to find the x-intercepts of \[ f(x) = x^2 - x - 2, \] factorize it as: \[ (x - 2)(x + 1) = 0. \] This gives:
- \( x = 2 \)
- \( x = -1. \)
y-intercept
The y-intercept of a quadratic function is the point where the graph crosses the y-axis. At this point, \( x = 0 \). To find the y-intercept, substitute \( x = 0 \) into the quadratic equation. For instance, in \[ f(x) = x^2 - x - 2, \] substituting \( x = 0 \) gives: \[ f(0) = 0^2 - 0 - 2 = -2. \] So, the y-intercept is \( (0, -2) \). The y-intercept is always the constant term \( c \) in the quadratic equation \( ax^2 + bx + c \), because this is what remains when \( x \) is zero.
domain and range
The domain and range are important characteristics of a quadratic function. The domain of any quadratic function is all real numbers, \[ (-\infty, \infty), \] as we can input any real number for \( x \) into the function. The range depends on the vertex and the direction of the parabola.
- If the parabola opens upwards (\( a > 0 \)), the range is from the y-coordinate of the vertex to infinity.
- If the parabola opens downwards (\( a < 0 \)), the range is from negative infinity to the y-coordinate of the vertex.
