Problem 5
Question
For Problems 1 through 8, graph the function. Label the \(x\) - and \(y\) -intercepts and the coordinates of the vertex. $$ f(x)=\left|-x^{2}+1\right| $$
Step-by-Step Solution
Verified Answer
The vertex of the function is at (0,1); the intercepts are at (-1,0) and (1,0). The graph is a 'W' shaped curve that corresponds to an absolute value quadratic function.
1Step 1: Solve for the Vertex
The vertex of the quadratic function occurs at \(x = -b / 2a\) where \(a, b, c\) are coefficients from the quadratic equation \(ax^2 + bx + c\). Here, \(a = -1\) and there is no \(b\) term, therefore \(x = 0\) is the x-coordinate of the vertex. To find the y-coordinate, replace \(x\) in the equation with the x-coordinate from earlier, therefore, \(f(0) = |-0^2 + 1| = 1\), hence the vertex is at (0,1).
2Step 2: Find X-intercepts and Y-intercept
The x-intercepts are found by solving the equation \(f(x)=0\), thus we have \(-x^2 + 1 = 0\) which gives \(x = -1, 1\) as the x-intercepts. For the y-intercept, set \(x = 0\) and solve for \(f(x)\). This gives the point (0,1) as also the y-intercept.
3Step 3: Graph the Quadratic Function
To graph this function, first mark the x-intercepts (-1, 0), (1, 0) and the vertex (0, 1). Given that this is an absolute value function with its vertex above the x-axis, the graph will look like a 'W' shape, reflecting the negative y values upwards. Connect the points to form a smooth curve. This forms the graph of the function.
Key Concepts
Vertex of a ParabolaX-interceptsY-interceptAbsolute Value Function
Vertex of a Parabola
The vertex of a parabola is a special point that indicates the maximum or minimum of the quadratic function, depending on its opening direction. For quadratic equations of the form \( ax^2 + bx + c \), the vertex can be calculated using the formula \( x = \frac{-b}{2a} \). In the exercise provided, the equation \( f(x) = |-x^2 + 1| \) corresponds to \( a = -1 \) and \( b = 0 \), therefore, the x-coordinate of the vertex is \( x = 0 \).
To discover the y-coordinate, substitute \( x = 0 \) back into the equation, resulting in \( f(0) = |-0^2 + 1| = 1 \). Thus, the vertex is at the point \((0, 1)\). This vertex being above the x-axis signifies the topping point of the "W" shape curve created by the absolute value function applied to the quadratic expression.
To discover the y-coordinate, substitute \( x = 0 \) back into the equation, resulting in \( f(0) = |-0^2 + 1| = 1 \). Thus, the vertex is at the point \((0, 1)\). This vertex being above the x-axis signifies the topping point of the "W" shape curve created by the absolute value function applied to the quadratic expression.
X-intercepts
X-intercepts are points where the graph of the function crosses the x-axis. In mathematical terms, these are the solutions of the equation \( f(x) = 0 \). For the equation given, \( f(x) = |-x^2 + 1| \), set the inside function to zero, \(-x^2 + 1 = 0 \).
After solving, you find that the x-intercepts occur at \( x = -1 \) and \( x = 1 \), making them \((-1, 0)\) and \((1, 0)\) respectively on the graph. These x-intercepts tell us where the directed graph reflects on the x-axis, creating the mirror-like 'W' pattern.
After solving, you find that the x-intercepts occur at \( x = -1 \) and \( x = 1 \), making them \((-1, 0)\) and \((1, 0)\) respectively on the graph. These x-intercepts tell us where the directed graph reflects on the x-axis, creating the mirror-like 'W' pattern.
Y-intercept
The y-intercept is the point where the graph crosses the y-axis, which occurs when \( x = 0 \). Finding this is quite simple: substitute \( x = 0 \) in the function \( f(x) = |-x^2 + 1| \).
Conducting this substitution yields \( f(0) = |0^2 + 1| = 1 \). The coordinate \((0, 1)\) represents the y-intercept of the graph. Interestingly, for the given function, this point is both the y-intercept and the vertex. This duality of roles emphasizes its significance in sketching the graph effectively.
Conducting this substitution yields \( f(0) = |0^2 + 1| = 1 \). The coordinate \((0, 1)\) represents the y-intercept of the graph. Interestingly, for the given function, this point is both the y-intercept and the vertex. This duality of roles emphasizes its significance in sketching the graph effectively.
Absolute Value Function
An absolute value function is a type of piecewise function that results in non-negative outputs. The absolute value is represented by the symbol \(|\cdot|\). For our function \( f(x) = |-x^2 + 1| \), applying the absolute value transforms the outputs less than zero into their positive counterparts.
The impact of the absolute value on the graph is visually distinctive: parts of the quadratic curve that would dip below the x-axis are flipped or "reflected" upward. This generates a 'W' shaped graph due to the nature of squaring the x-term and applying negation and absolute value.
The impact of the absolute value on the graph is visually distinctive: parts of the quadratic curve that would dip below the x-axis are flipped or "reflected" upward. This generates a 'W' shaped graph due to the nature of squaring the x-term and applying negation and absolute value.
- The downward shape from \(-x^2\) becomes upward upon applying absolute value.
- Parts of the curve that would normally enter negative y-values remain above the axis.
Other exercises in this chapter
Problem 4
We know that Revenue \(=\) (price) . (quantity). Suppose a certain company has a monopoly on a good. If the company wants to increase its revenue it can do so b
View solution Problem 4
For each of the quadratics, identify the \(x\) - and \(y\) -coordinates of the vertex and determine whether the vertex is the highest point on the curve or the
View solution Problem 5
Refer to Problem 4 for your answers to this question. (a) How many solutions are there to \(2(x-3)^{2}-5=-6 ?\) (b) How many solutions are there to \(-4(x+1)^{2
View solution Problem 5
Suppose that \(q\), the quantity of gas (in gallons) demanded for heating purposes, is given by \(q=m p+b\), where \(m\) and \(b\) are constants \((m\) negative
View solution