Vertical and horizontal shifts

Vertical and horizontal shifts • If f is the function y = f(x) = x2, then we can plot points and draw its graph as: • If we add 1 (outside change) to f(x), we have y = f(x) + 1 = x2 + 1. We simply take the graph above and move it up 1 unit to get the new graph. y x y (2,5) x

If we replace x by x+2 (inside change) to form the function y = f(x+2) = (x+2)2, then the corresponding graph is obtained from the graph of y = x2 by moving it 2 units to the left along the x-axis. • If we replace x by x–1 (inside change) to form the function y = f(x–1) = (x–1)2, then the corresponding graph is obtained from the graph of y = x2 by moving it 1 unit to the right along the x-axis. y x (-2,0) y x (1,0)

If y = g(x) is a function and k is a constant, then the graph of: • y = g(x) + k is the graph of y = g(x) shifted vertically by |k| units. If k > 0, the shift is up, and if k < 0, the shift is down. • y = g(x+k) is the graph of y = g(x) shifted horizontally by |k| units. If k > 0, the shift is left, and if k < 0, the shift is right. Horizontal and vertical shifts of the graph of a function are called translations.

An example which combines horizontal and vertical shifts • Problem. Use the graph of y = f(x) = x2 to sketch the graph of g(x) = f(x–2) – 1 = (x–2)2 – 1. Solution. The graph of g is the graph of f shifted to the right by 2 units and down 1 unit as shown below. y x (2,-1)

Reflections and symmetry • Suppose that we are given the function y = f(x) as shown. • If we define y = g(x) = –f(x), then the graph of g may be obtained by reflecting the graph of f vertically across the x-axis as shown next. y x y x

Continuation of example from previous slide • If we define y = h(x) = f(–x), then the graph of h is obtained by reflecting the graph of f horizontally across the y-axis as shown next. • Next, we define y = p(x) = –f(–x). The graph of p is obtained by reflecting the graph of f about the origin as shown next. y x y x

For any function f: • The graph of y = –f(x) is the reflection of the graph of y = f(x) across the x-axis. • The graph of y = f(–x) is the reflection of the graph of y = f(x) across the y-axis. • The graph of y = –f(–x) is the reflection of the graph of y = f(x) about the origin. Note that this reflection can be obtained by applying the two previous reflections in sequence.

Symmetries of graphs • A function is called an even function if, for all values of x in the domain of f, The graph of an even function is symmetric across the y-axis. Examples of even functions are power functions with even exponents, such as y = x2, y = x4, y = x6, ... • A function is called an odd function if, for all values of x in the domain of f, The graph of an odd function is symmetric about the origin. Examples of odd functions are power functions with odd exponents, such as y = x1, y = x3, y = x5, ...

Problem. Is the function f(x) = x3+x even, odd, or neither? Solution. Since –2 = f(–1) is not equal to f(1) = 2, it follows that f is not even. Since f(–x) = = –f(x), it follows that f is odd. y = x3+x Note the symmetry about the origin.

Problem. Is the function f(x) = |x| even, odd, or neither? Solution. Since f(–x) = |x| = f(x), it follows that f is even. Since 1 = f(–1) is not equal to –f(1) = –1, it follows that f is not odd. • Question. Is it possible for a function to be both even and odd? y = |x| Note the symmetry about the y-axis.

Combining shifts and reflections--an example • In an earlier example, we discussed an investment of $10000 in the latest dotcom venture. This investment had a value of 10000(0.95)t dollars after t years. Suppose that we want to graph the amount of the loss after t years for this investment. The formula for the loss is: 10000 – 10000(0.95)t • The loss is graphed on the next slide using Maple. Shift Upwards Reflect across t-axis

Use of Maple to graph loss on dotcom investment > plot({10000,10000-10000*(0.95)^t},t=0..80, color=black,labels=["t","L"]); The graph of the loss has a horizontal asymptote, L = 10000.

Vertical Stretches and Compressions • If f(x) = x2 and g(x) = 5x2, then the graph of g is obtained from the graph of f by stretching it vertically by a factor of 5 as the following Maple plot shows:

If f(x) = x2 and g(x) = -5x2, then the graph of g is obtained from the graph of f by stretching it vertically by a factor of 5 and then reflecting it across the x-axis as the following Maple plot shows:

If we compare the graphs of f(x) = x2 and g(x) = (1/2)x2, we notice that the graph of g can be found by vertically compressing the graph of f by a factor of 1/2. • Generalizing the above examples yields the following: If f is a function and k is a constant, then the graph of y = kf(x) is the graph of y = f(x) • Vertically stretched by a factor of k, if k > 1. • Vertically compressed by a factor of k, if 0 < k < 1. • Vertically stretched or compressed by a factor |k| and reflected across the x-axis, if k < 0.

Vertical Stretch Factors and Average Rates of Change • If f(x) = x2 and g(x) = 5x2, we compute the average rates of change of the two functions on the interval [1,3] as follows: • The above computation illustrates a general fact:

If f(x) = 4–x2 and g(x) = 4 – (2x)2, then the graph of g is obtained from the graph of f by compressing it horizontally by a factor of 1/2 as the following Maple plot shows:

If f(x) = 4–x2 and g(x) = 4 – (0.5x)2, then the graph of g is obtained from the graph of f by stretching it horizontally by a factor of 2 as the following Maple plot shows:

Generalizing the two previous examples yields the following results for horizontal stretch or compression. • If f is a function and k is a positive constant, then the graph of y = f(kx) is the graph of f • Horizontally compressed by a factor of 1/k if k > 1. • Horizontally stretched by a factor of 1/k if k < 1. If k < 0, then the graph of y = f(kx) also involves a horizontal reflection about the y-axis.

The Family of Quadratic Functions • A quadratic function is a function with a formula in one of the following forms: • Standard form: y = ax2+bx+c, where a, b, c, are constants, • Vertex form: y = a(x–h)2+k, where a, h, k are constants, • The graph of a quadratic function is called a parabola. • Conversion from one form to the other for a quadratic function is discussed on the next slide.

To convert a quadratic function from vertex form to standard form, simply multiply out the squared term. To convert from standard form to vertex form, we complete the square as illustrated in the following example. • Example. Put the following quadratic function into vertex form by completing the square. (1) Factor out the coefficient of x2, which is – 4. (2) Add and subtract the square of half the coefficient of the x-term. (3) Write the equation in vertex form. perfect square

The Vertex of a Parabola • Recall that the graph of a quadratic is called a parabola. • The parabola corresponding to y = a(x–h)2+k: • Has vertex (h, k). • Has axis of symmetry x = h. • Opens upward if a > 0 or downward if a < 0.

Finding the vertex of a parabola • Example. For the previous example, graph the parabola and find the vertex. We note that the vertex is at (–3/2, 1). The graph follows:

Finding a formula for a parabola • If we know the vertex of a quadratic function and one other point, we can use the vertex form to find its formula, as shown in the following example. • Example. A parabola has vertex at (–3, 2) and (0, 5) is on the parabola. Find the formula for the corresponding quadratic, f(x). Use the vertex form with h = –3 and k =2. This results in To find the value of a, we substitute x = 0 and y = 5 into this formula, obtaining a = 1/3. The formula is therefore

Finding a formula for a parabola, continued. • If three points on a parabola are given, we can use the standard form of the corresponding quadratic to find the formula. • Example. Suppose the points (0, 6), (1, 0), and (3, 0) are on a parabola. Find a formula for the parabola. Use the standard form: y = ax2+bx+c. Since (0, 6) is on the parabola, it follows that c = 6. From the other two points, we have: This system can be solved simultaneously for a and b. We obtain a = 2 and b = –8. Thus, the equation of the parabola is y = 2x2–8x+6.

Finding the zeros of a quadratic function. • The zeros of a function f are values of x for which f(x) = 0. • In addition to the standard and vertex forms, some quadratic functions f(x) can also be expressed in factored form: • Example. Find the zeros of f(x) = x2–x –6. Set f(x) = 0 and solve for x. We have x2–x –6 = 0. We next express f(x) in factored form, so it will be easy to find the zeros. The zeros are x = 3 and x = –2. Note that these are the values r and s from the factored form.

Finding a formula for a parabola using the factored form • Example. Suppose the points (0, 6), (1, 0), and (3, 0) are on a parabola, as in a previous example. Find a formula for the parabola using the factored form. Since the parabola has x-intercepts at x = 1 and x = 3, its formula is Substituting x = 0, y = 6 gives 6 = 3a or a = 2. Thus, the equation is: If we multiply this out, we get y = 2x2–8x+6, which is the same result as before.

Two methods for finding the zeros of a quadratic • The first method involves completing the square. Suppose we want the roots of x2 + 3x + 2 = 0. If we complete the square as before, we get (x + 1.5)2– 0.25 = 0. If we rewrite this as (x + 1.5)2 = 0.25, we can take the square root of both sides of the equation to get x + 1.5 = 0.5, which gives x = –1 and x = –2. • The other method involves the use of the quadratic formula, which was presented in a previous slide lecture. If we apply the quadratic formula to x2 + 3x + 2 = 0, we get This reduces to x = –1.5 0.5 . Again, x = –1 and x = –2.

What does it mean if a quadratic does not have real zeros? It means that the graph of the corresponding parabola does not cross the x-axis. • Problem. If we have 4 feet of string, what is the rectangle of largest area which we can enclose with the string? Solution. If we let one side of the rectangle have length x, then the other side must have length (4–2x)/2. That is, the other side is 2–x. Therefore, the area of the rectangle is a(x) = x(2–x) = –x2+2x. If we write this in vertex form, we have a(x) = –(x–1)2+1. Thus, the vertex is at (1, 1), and the rectangle of maximum area is a square with a side length of 1 foot.

Summary for Transformation of Functions and their Graphs • If y = g(x) is a function and k is a constant, then the graph of y = g(x) + k is the graph of y = g(x) shifted vertically by |k| units. • If y = g(x) is a function and k is a constant, then the graph of y = g(x+k) is the graph of y = g(x) shifted horizontally by |k| units. • A function is called an even function if, for all values of x in the domain of f, f(–x) = f(x). The graph of an even function is symmetric across the y-axis. • A function is called an odd function if, for all values of x in the domain of f, f(–x) = –f(x). The graph of an odd function is symmetric about the origin.

Summary for Transformation of Fcts and their Graphs, cont’d • When a function f(x) is replaced by kf(x), the graph is vertically stretched or compressed and the average rate of change on any interval is also multiplied by k. If k is negative, a vertical reflection about the x-axis is also involved. • When a function f(x) is replaced by f(kx), the graph is horizontally stretched or compressed by a factor of 1/|k| and, if k < 0, reflected horizontally about the y-axis. • A quadratic function has a formula in either standard form or vertex form. Completing the square converts standard to vertex form. Vertex form is used to find the max or min value of the quadratic function. A quadratic function has 0, 1, or 2 real zeros. If a quadratic function has real zeros, it can also be represented in factored form. Methods for finding the formula for a quadratic function from given data points were discussed.

Vertical and horizontal shifts