Problem 40
Question
Use transformations of graphs to sketch a graph of \(y=f(x)\) by hand. $$f(x)=\frac{1}{2}(x+2)^{2}$$
Step-by-Step Solution
Verified Answer
The graph is a parabola opening upwards, shifted 2 units left, and vertically compressed by a factor of 2.
1Step 1: Identify the Basic Graph
The given function is a transformation of the basic quadratic function \(y = x^2\). Identify that the basic graph shape of \(y = x^2\) is a parabola opening upwards with its vertex at the origin \((0,0)\).
2Step 2: Apply Horizontal Shift
The term \((x+2)\) indicates a horizontal shift. Specifically, \(x\) is replaced with \(x+2\), which shifts the graph 2 units to the left. So, the new vertex of the parabola is now at \((-2,0)\).
3Step 3: Apply Vertical Stretch and Compression
The factor \(\frac{1}{2}\) in front of \((x+2)^2\) is a vertical compression. This means the graph of the parabola becomes wider since any \(y\)-value for a given \(x\) is now half of what it would be in \(y=(x+2)^2\).
4Step 4: Sketch the Transformed Graph
Start by marking the new vertex at \((-2,0)\). Since the graph opens upwards and is vertically compressed, sketch the parabola with points that reflect these transformations, ensuring that it is symmetric around the vertical line \(x=-2\).
Key Concepts
Quadratic FunctionsParabolasHorizontal ShiftVertical Compression
Quadratic Functions
Quadratic functions are fundamental mathematical expressions that take the form \( y = ax^2 + bx + c \). These functions represent a parabola when graphed on the coordinate plane. Their most basic form, \( y = x^2 \), creates a symmetrical curve that opens upwards. The shape and orientation of this curve can change depending on the values of the coefficients \( a \), \( b \), and \( c \). In the formula, \( a \) determines the direction and width of the parabola. If \( a \) is positive, the parabola opens upwards; if negative, it opens downwards. A larger value of \( a \) results in a narrower parabola, whereas a smaller \( a \) leads to a wider curve. This makes quadratic functions incredibly versatile in graph transformations.
Parabolas
Parabolas are U-shaped curves that represent the graph of a quadratic function. They have a distinct axis of symmetry, a vertical line that divides the parabola into two mirror-like halves. The vertex, located on the axis of symmetry, is a critical point where the parabola changes direction. The standard parabola \( y = x^2 \) has its vertex at \( (0, 0) \). This point is also the minimum point if the parabola opens upwards and the maximum if it opens downwards.
- The direction in which the parabola opens depends on the sign of the leading coefficient \( a \).
- The vertex is a vital reference in graph transformations, serving as a starting point for shifts and stretches.
Horizontal Shift
A horizontal shift in a graph involves moving the entire graph left or right on the coordinate plane without altering its shape. For a quadratic function \( y = (x-h)^2 \), the term \( h \) dictates the horizontal shift. If \( x \) is replaced with \( x + 2 \), as in the exercise \( y = \frac{1}{2}(x+2)^2 \), the graph shifts 2 units to the left. This change in position redefines the location of the vertex from \( (0, 0) \) to \( (-2, 0) \). During a horizontal shift,
- The symmetry and orientation of the parabola remain unchanged.
- The axis of symmetry moves along with the vertex, helping to maintain the parabola's symmetry.
Vertical Compression
Vertical compression involves altering the vertical scale of a graph, effectively compressing the shape without changing its position on the horizontal axis. This type of transformation is indicated by a coefficient multiplying the function. In the expression \( y = \frac{1}{2}(x+2)^2 \), the \( \frac{1}{2} \) compresses the parabola vertically, making the graph appear wider compared to the standard \( y = x^2 \).
- A vertical stretch would occur if the multiplying factor were greater than 1, making the parabola narrower.
- The vertex remains in place during vertical compression, ensuring the graph's minimum or maximum point stays constant.
Other exercises in this chapter
Problem 40
Graph each equation by hand. $$y=4-8 x, y=|4-8 x|$$
View solution Problem 40
Without a graphing calculator, determine the domain and range of the functions. $$f(x)=\sqrt{x+1}-10$$
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Describe how the graph of the given function can be obtained from the graph of \(y=[x]\). $$y=[x]-1.5$$
View solution Problem 41
Without a graphing calculator, determine the domain and range of the functions. $$f(x)=(x-1)^{3}+4$$
View solution