Problem 64
Question
Begin by graphing the standard quadratic function, \(f(x)-x^{2} .\) Then use transformations of this graph to graph the given function. $$ h(x)-\frac 12(x-1)^{2}-1 $$
Step-by-Step Solution
Verified Answer
The graph of the function \(h(x) = \frac{1}{2}(x-1)^{2} - 1\) can be obtained from the graph of the standard function \(f(x) = x^{2}\) by shifting 1 to the right, then 1 unit down and finally applying a vertical compression by a factor of \frac{1}{2}.
1Step 1: Graph the Standard Function
Start by drawing the graph of the function f(x) = x^{2}. This is a standard quadratic function that forms a parabola that opens upwards, with the vertex (the lowest point) at the origin (0,0).
2Step 2: Identify the Transformations
Next, identify the transformations from the given function h(x) = \frac{1}{2}(x-1)^2 - 1. Here there are three transformations: a horizontal shift to the right by 1 unit (from the -1 inside the bracket), a vertical shift downward by 1 unit (from the -1 outside the bracket), and a vertical compression by a factor of a \frac{1}{2} (from the coefficient of x^2, which is \frac{1}{2}).
3Step 3: Apply the Transformations
Apply the transformations to the graph of f(x) one by one. First, shift the graph 1 unit to the right (this is the effect of (x-1)). This will move the vertex of the parabola from (0,0) to (1,0). Then, shift the graph 1 unit down (this is the effect of -1). This moves the vertex down to the point (1,-1). Lastly, perform the vertical compression by a factor of \frac{1}{2}, which makes the parabola 'wider' or more 'opened out'. The new graph now represents the function h(x).
Key Concepts
Graph TransformationsParabolasVertex Form
Graph Transformations
Graph transformations are like a roadmap that helps you change one graph into another. Imagine you have the basic graph of a quadratic function, which looks like a U-shaped curve, called a parabola. By applying transformations, you can move, stretch, or squash this curve without changing its fundamental shape.
In our problem, the original function is a simple parabola: \( f(x) = x^2 \). The transformed function is \( h(x) = \frac{1}{2}(x-1)^2 - 1 \). Let's break down what happens to the graph of \( f(x) \) with these transformations:
In our problem, the original function is a simple parabola: \( f(x) = x^2 \). The transformed function is \( h(x) = \frac{1}{2}(x-1)^2 - 1 \). Let's break down what happens to the graph of \( f(x) \) with these transformations:
- **Horizontal Shift**: Moving the graph left or right. \( (x-1) \) shifts the graph to the right by 1 unit.
- **Vertical Shift**: Raising or lowering the graph. Here, the \(-1\) outside the parentheses shifts it downward by 1 unit.
- **Vertical Compression**: Making the graph wider or narrower. The coefficient \( \frac{1}{2} \) compresses it, making the parabola wider.
Parabolas
Parabolas are elegant curves that you often see in quadratic functions. They resemble the shape of a U or an inverted U, depending on the direction they open. The most basic form is defined by \( f(x) = x^2 \), which creates a parabola opening upwards.
A parabola has some key features:
A parabola has some key features:
- **Vertex**: The highest or lowest point of the parabola. For \( f(x) = x^2 \), the vertex is at the origin (0,0).
- **Axis of Symmetry**: A vertical line that divides the parabola into two mirror-image halves. For the function \( f(x) = x^2 \), this line is \( x = 0 \).
- **Direction**: Parabolas can open upwards or downwards. An upward opening happens when the coefficient of \( x^2 \) is positive, like in \( f(x) = x^2 \). If it’s negative, the parabola opens downwards.
Vertex Form
The vertex form of a quadratic equation is a useful tool for graphing parabolas efficiently. It's written as \( y = a(x-h)^2 + k \), where \((h, k)\) is the vertex of the parabola, and \(a\) dictates the parabola's width and direction.
In our function \( h(x) = \frac{1}{2}(x-1)^{2} - 1 \), comparing this to the vertex form helps us easily identify:
In our function \( h(x) = \frac{1}{2}(x-1)^{2} - 1 \), comparing this to the vertex form helps us easily identify:
- **Vertex**: \((1, -1)\), showing where the parabola's peak or trough lies.
- **"a" Value**: \(a = \frac{1}{2}\), indicating that the parabola is wider than normal because the value is less than 1. If \(a\) were negative, the parabola would open downwards.
Other exercises in this chapter
Problem 63
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Complete the square and write the equation in standard form. Then give the center and radius of each circle and graph the equation. $$ x^{2}+y^{2}+3 x+5 y+\frac
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Find and simplify the difference quotient $$\frac{f(x+h)-f(x)}{h}, h \neq 0$$ for the given function. $$f(x)=3 x^{2}+x+5$$
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