Chapter 2: Problem 5
Sketch the graph of the given function \(f\). $$ f(x)=-2 x^{2}+1 $$
Short Answer
Expert verified
The graph is a downward-opening parabola with vertex (0,1) and y-intercept 1.
Step by step solution
01
Identify the Type of Function
Recognize that the function given is a quadratic function, specifically written in the standard form of a parabola, which is \(f(x) = ax^2 + bx + c\). In this case, \(a = -2\), \(b = 0\), and \(c = 1\).
02
Determine the Vertex
To find the vertex of the parabola, use the formula \(x = -\frac{b}{2a}\). Here, since \(b = 0\), the vertex's \(x\)-coordinate is 0. Plug \(x = 0\) back into \(f(x)\) to get \(-2(0)^2 + 1 = 1\). Thus, the vertex of the parabola is at \((0, 1)\).
03
Determine the Axis of Symmetry
The axis of symmetry for a quadratic function in standard form is always \(x = -\frac{b}{2a}\). With \(b = 0\), the axis of symmetry is \(x = 0\).
04
Identify the Direction of the Parabola
The sign of the coefficient \(a\) determines the direction of the parabola. If \(a\) is negative, the parabola opens downwards. Here, \(a = -2\), so the parabola opens downwards.
05
Find the Y-Intercept
The y-intercept occurs when \(x = 0\). Substituting \(x = 0\) into the function gives \(f(0) = -2(0)^2 + 1 = 1\). Thus, the y-intercept is the same as the vertex, \((0, 1)\).
06
Calculate Another Point
Choose another x-value, say \(x = 1\), to find another point on the graph. Substitute \(x = 1\) in the function: \(f(1) = -2(1)^2 + 1 = -1\). So, the point \((1, -1)\) lies on the graph. Similarly, by symmetry, \((-1, -1)\) will also lie on the graph.
07
Sketch the Graph
Using the vertex \((0, 1)\), the axis of symmetry \(x = 0\), the direction of the parabola, and points \((1, -1)\) and \((-1, -1)\), sketch the downward opening parabola. Ensure the parabola is symmetric with respect to the y-axis.
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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 with a degree of 2. It is usually expressed in the standard form as \( f(x) = ax^2 + bx + c \). The graph of a quadratic function is a curve called a parabola. The values \( a \), \( b \), and \( c \) are coefficients that determine the shape and position of the parabola in a coordinate plane.
For the given function, \( f(x) = -2x^2 + 1 \), the coefficients are \( a = -2 \), \( b = 0 \), and \( c = 1 \).
This specific format shows that our parabola will be vertically in place. If the value of \( a \) were multiplied by a negative number, our parabola opens downward. Conversely, if \( a \) were positive, it would opened upward.
For the given function, \( f(x) = -2x^2 + 1 \), the coefficients are \( a = -2 \), \( b = 0 \), and \( c = 1 \).
This specific format shows that our parabola will be vertically in place. If the value of \( a \) were multiplied by a negative number, our parabola opens downward. Conversely, if \( a \) were positive, it would opened upward.
- \( a \) affects the width and the direction of the parabola.
- When \( b \) is zero, the vertex is on the y-axis.
- The value of \( c \) represents the y-intercept.
Vertex
The vertex of a parabola is its highest or lowest point, depending on its opening direction. For quadratic functions in standard form, you can find the vertex using the formula: \( x = -\frac{b}{2a} \). This formula gives the x-coordinate of the vertex. Once you have the x-coordinate, substitute it back into the function to find the y-coordinate.
For \( f(x) = -2x^2 + 1 \), with \( b = 0 \), the calculation simplifies to \( x = 0 \). Plug \( x = 0 \) back into the function to get the y-coordinate, resulting in \( f(0) = 1 \).
Therefore, the vertex of this parabola is \((0, 1)\). Because \( a = -2 \) is negative, this vertex is the maximum point of the parabola. It appears at the top of our downward opening curve.
For \( f(x) = -2x^2 + 1 \), with \( b = 0 \), the calculation simplifies to \( x = 0 \). Plug \( x = 0 \) back into the function to get the y-coordinate, resulting in \( f(0) = 1 \).
Therefore, the vertex of this parabola is \((0, 1)\). Because \( a = -2 \) is negative, this vertex is the maximum point of the parabola. It appears at the top of our downward opening curve.
Axis of Symmetry
The axis of symmetry is a vertical line that splits the parabola into two mirror-image halves. This is fundamental in understanding symmetry in quadratic functions as it provides a reference line for sketching the parabola.
The equation for the axis of symmetry in a quadratic function in standard form is \( x = -\frac{b}{2a} \). Similar to finding the vertex, for our function \( f(x) = -2x^2 + 1 \), with \( b = 0 \), the axis of symmetry is \( x = 0 \).
The equation for the axis of symmetry in a quadratic function in standard form is \( x = -\frac{b}{2a} \). Similar to finding the vertex, for our function \( f(x) = -2x^2 + 1 \), with \( b = 0 \), the axis of symmetry is \( x = 0 \).
- This tells us that the parabola is symmetric about the line \( x = 0 \).
- Every point on the parabola one side of this line has an exactly mirrored point on the opposite side.
Y-Intercept
The y-intercept of a graph is where the graph intersects the y-axis. For quadratic functions, it exists when \( x = 0 \). At this point, the function simplifies to \( f(x) = c \) because all \( x \)-terms become zero.
For \( f(x) = -2x^2 + 1 \), the y-intercept is obtained by setting \( x = 0 \). The resulting value is \( f(0) = 1 \).
For \( f(x) = -2x^2 + 1 \), the y-intercept is obtained by setting \( x = 0 \). The resulting value is \( f(0) = 1 \).
- Therefore, the y-intercept of our function is at the point \((0, 1)\), which is also the vertex in this case.
- This point gives us an essential reference for sketching the parabola as it provides a specific starting point.