Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 0: Problem 35
Graph each function. $$ f(x)=6-x^{2} $$
Short Answer
Expert verified
The graph is a downward opening parabola with vertex (0, 6), y-intercept (0, 6), and x-intercepts (卤鈭�6, 0).
Step by step solution
01
Identify the type of function
The function given is a quadratic in the form of \( f(x) = a - bx^2 \) which represents a downward opening parabola because the coefficient of \( x^2 \) is negative. Here, \( a = 6 \) and \( b = 1 \).
02
Find the vertex of the parabola
In the standard quadratic function \( f(x) = a - b(x - h)^2 + k \), the vertex is \((h, k)\). For \( f(x) = 6 - x^2 \), it can be written as \( f(x) = -1(x - 0)^2 + 6 \). From this, we identify the vertex is \((0, 6)\).
03
Determine the y-intercept
The y-intercept is the point where the graph crosses the y-axis. To find it, set \( x = 0 \). Thus, \( f(0) = 6 - 0^2 = 6 \). Therefore, the y-intercept is \((0, 6)\).
04
Find x-intercepts
The x-intercepts occur where the function equals zero: \( 6 - x^2 = 0 \). Solving \( x^2 = 6 \), we get \( x = \pm\sqrt{6} \). Thus, the x-intercepts are \((\sqrt{6}, 0)\) and \((-\sqrt{6}, 0)\).
05
Sketch the graph
Using the points and vertex found, sketch the parabola starting from the vertex \((0, 6)\) and passing through the points \((0, 6)\), \((\sqrt{6}, 0)\), and \((-\sqrt{6}, 0)\). Ensure the parabola opens downward based on the coefficient \(-1\) of \(x^2\).
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 Functions
A quadratic function is a specific type of polynomial function that can be represented in the general form: \( f(x) = ax^2 + bx + c \). Here, \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). One distinctive feature of quadratic functions is their graph. The graph is a symmetrical curve referred to as a parabola. Key characteristics of a parabola include:
- It can either open upwards or downwards. This is determined by the sign of the coefficient \( a \). When \( a > 0 \), the parabola opens upwards. Conversely, when \( a < 0 \), it opens downwards.
- It has a single vertex, which represents either the maximum or minimum point of the parabola.
- The parabola is symmetrical around a vertical line known as the axis of symmetry.
Vertex Form
The vertex form of a quadratic function is particularly helpful for quickly finding the vertex of the parabola. It is written as:\[ f(x) = a(x - h)^2 + k \]In this format:
- \( (h, k) \) is the vertex of the parabola.
- \( a \) determines the direction and steepness of the parabola.
- The parabola opens upwards if \( a > 0 \) and downwards if \( a < 0 \).
X-intercepts
The x-intercepts, also known as roots or zeros, of a quadratic function are points where the parabola crosses the x-axis. These points occur where the function equals zero. For any quadratic equation, solving for the x-intercepts involves setting \( f(x) = 0 \) and finding the values of \( x \):For our function \( f(x) = 6 - x^2 \), set it to zero, producing:\[ 6 - x^2 = 0 \]Solving this yields:\[ x^2 = 6 \]The solutions are \( x = \pm\sqrt{6} \). Hence, we can say the x-intercepts are the points \( (\sqrt{6}, 0) \) and \( (-\sqrt{6}, 0) \). These intercepts are where the graph touches the x-axis and are crucial points in sketching the parabola.
Y-intercept
The y-intercept of a quadratic function is where the graph intersects the y-axis. It occurs when the input \( x \) is set to zero. This point is essential for sketching the parabola, as it provides a starting point on the vertical axis.In our function \( f(x) = 6 - x^2 \), the y-intercept is calculated by substituting \( x = 0 \) into the function:\[ f(0) = 6 - 0^2 = 6 \]Therefore, the y-intercept is the point \( (0, 6) \). This tells us that the graph passes through this point on the y-axis. Having both the y-intercept and the vertex provides us key locations that aid in accurately sketching the graph of the function.
