Problem 22
Question
Sketch the graphs of the following functions. Beneath the sketch of the function, sketch the graph of the derivative. If the graph is the graph of a quadratic, label the coordinates of the vertex of the corresponding parabola; if the graph has corners, label the coordinates of the corners. (a) \(f(x)=-(x+2)^{2}\) (b) \(f(x)=(x-2)^{2}+3\) (c) \(f(x)=(x-2)(x+4)\) (d) \(f(x)=-2(x-2)(x+4)\) (e) \(f(x)=|x+4|+2\) (f) \(f(x)=|2 x|-3\)
Step-by-Step Solution
Verified Answer
The vertex coordinates of the parabolas for \(f(x)=-(x+2)^{2}\), \(f(x)=(x-2)^{2}+3\), \(f(x)=(x-2)(x+4)\), and \(f(x)=-2(x-2)(x+4)\) are (-2,0), (2,3), (1,-2), and (1,-8) respectively. The corners for \(f(x)=|x+4|+2\) and \(f(x)=|2 x|-3\) are at (-4,2) and (0,-3) respectively.
1Step 1: Graph the Functions and Identify the Characteristics
We begin with the quadratic functions. Quadratic functions have a general form of \(f(x)=a(x-h)^{2}+k\), where (h,k) is the vertex of the parabola. If the coefficient 'a' is positive, the parabola is U-shaped and if it's negative, the parabola is inverted. For \(f(x)=-(x+2)^{2}\), plot the graph and identify the vertex at (-2,0). For \(f(x)=(x-2)^{2}+3\), do the same and identify the vertex at (2,3). For \(f(x)=(x-2)(x+4)\) and \(f(x)=-2(x-2)(x+4)\), expand the polynomial, set the derivative equal to zero, solve for x, and find the corresponding y values to identify the vertex and sketch the graph. As for the absolute functions, \(f(x)=|x+4|+2\) and \(f(x)=|2 x|-3\), draw a V-shaped graph. For \(f(x)=|x+4|+2\), shift 4 units to the left and 2 units up, and for \(f(x)=|2 x|-3\), stretch the graph by a factor of 2 and shift down by 3 units.
2Step 2: Compute the Derivatives and Graph their Functions
The derivative of a function gives the rate of change of the function at a specific point. For the quadratic functions, compute the derivative using the power rule, \(d/dx[x^n] = nx^{n-1}\), and the chain rule, \(d/dx[f(g(x))] = f'(g(x))g'(x)\). For the absolute value functions, we have to split them at the corner and compute the derivatives of the respective linear functions separately. Once the derivative functions are computed, sketch their graphs.
3Step 3: Label the Coordinates of the Vertex and Corners
For the quadratic functions, their vertex is labelled. For the absolute value functions, their corners are the points where the function changes direction. For \(f(x)=|x+4|+2\), we see a corner at (-4,2), and for \(f(x)=|2 x|-3\), the corner is at (0,-3). Label these points on the functions' graphs, and for the derivative functions, demonstrate a break at these points by indicating an open circle.
Key Concepts
Quadratic FunctionsDerivative CalculationsAbsolute Value FunctionsVertex IdentificationGraph Sketching
Quadratic Functions
Understanding quadratic functions is essential in calculus graphing because they form the simplest polynomial function with interesting properties like vertex and symmetry. These functions take the form \( f(x) = a(x-h)^2 + k \). The constants \( h \) and \( k \) determine the vertex's position, a crucial point in parabolas.
- When \( a > 0 \), the parabola opens upwards, resembling a U shape.
- When \( a < 0 \), it opens downwards, forming an inverted U shape.
Derivative Calculations
Calculating the derivative of a function is a way of finding its rate of change or the slope of the tangent line at any given point. For quadratic functions, the derivative is determined using basic calculus rules:
- Power Rule: \( \frac{d}{dx}[x^n] = nx^{n-1} \)
- Chain Rule: \( \frac{d}{dx}[f(g(x))] = f'(g(x))g'(x) \)
Absolute Value Functions
Absolute value functions involve expressions where the absolute value operator \(| \,|\) affects how they behave on a graph, resulting in a V-shaped graph. For example, the function \( f(x)=|x+4|+2 \) produces a sharp corner where the direction of the line changes.
- The basic transformation includes translating and scaling the graph.
- For \( f(x)=|x+4|+2 \), the graph shifts left by 4 units and up by 2 units.
- Similarly, \( f(x)=|2x|-3 \) means a vertical stretch by a factor of 2 and a downward shift by 3 units.
Vertex Identification
Identifying the vertex is vital in understanding quadratic functions because it represents the highest or lowest point of a parabola. It acts as a guide for graphing and finding function properties such as the axis of symmetry.
- In \( f(x)=a(x-h)^2+k \), the vertex is always \((h,k)\).
- It determines the shape and symmetry of a parabola.
- In graphs, it's the "peak" or "valley." For instance, the vertex of \( f(x)=(x-2)^2+3 \) is \((2,3)\).
Graph Sketching
Sketching graphs is an essential skill in understanding the behavior of functions visually. It lets us see the relationships between different aspects like the rate of change or position of vertices.
- Always start by identifying key points such as vertices for parabolas or corners for absolute value functions.
- For quadratic functions, knowing the vertex helps in determining the symmetry and direction of the parabola.
- The derivative's graph shows the slope at different points, important for understanding the change of rate in the function.
Other exercises in this chapter
Problem 20
In Problems 17 through 21, nd the equation of the parabola with the speci cations given. Vertex at \((1,5) ; y\) -intercept of 1
View solution Problem 21
In Problems 17 through 21, nd the equation of the parabola with the speci cations given. Vertex at \((-2,3)\); passing through \((1,-1)\)
View solution Problem 23
\text { Find the equation of the parabola through the points }(0,3),(1,0), \text { and }(2,-1) \text { . }
View solution Problem 24
Find the coordinates of the vertex of the parabola passing through the points \((2,0)\), \((-1,9)\), and \((1,-5)\) Decide upon a strategy for doing this proble
View solution