Introduction Transformations can be made to functions in two distinct ways: by transforming the core variable of the function (multiplying the independent variable by k), and by transforming the function as a whole (multiplying the dependent variable by k). Previously, we saw how adding some constant k to the variable of a function or to the entire function affected the graph of the function. In this lesson, we will see how multiplying by a constant affects the graph of a function. Given f(x) and a constant k, we will observe the transformations f(k x) and k f(x) : Replacing f(x) with k f(x) and f(k x)

Key Concepts Graphing and Points of Interest In the graph of a function, there are key points of interest that define the graph and represent the characteristics of the function. When a function is transformed, the key points of the graph define the transformation. The key points in the graph of a quadratic equation are the vertex and the roots, or x-intercepts : Replacing f(x) with k f(x) and f(k x)

Key Concepts, continued Multiplying the Dependent Variable by a Constant, k: k f(x) In general, multiplying a function by a constant will stretch or shrink (compress) the graph of f vertically. If k > 1, the graph of f(x) will stretch vertically by a factor of k (so the parabola will appear narrower). A vertical stretch pulls the parabola and stretches it away from the x-axis. If 0 < k < 1, the graph of f(x) will shrink or compress vertically by a factor of k (so the parabola will appear wider) : Replacing f(x) with k f(x) and f(k x)

Key Concepts, continued A vertical compression squeezes the parabola toward the x-axis. If k < 0, the parabola will be first stretched or compressed and then reflected over the x-axis. The x-intercepts (roots) will remain the same, as will the x-coordinate of the vertex (the axis of symmetry). While k f(x) = f(k x) can be true, generally k f(x) ≠ f(k x) : Replacing f(x) with k f(x) and f(k x)

Key Concepts, continued : Replacing f(x) with k f(x) and f(k x) Vertical stretches: when k > 1 in k f(x) The graph is stretched vertically by a factor of k. The x-coordinate of the vertex remains the same. The y-coordinate of the vertex changes. The x-intercepts remain the same.

Key Concepts, continued : Replacing f(x) with k f(x) and f(k x) Vertical compressions: when 0 < k < 1 in k f(x) The graph is compressed vertically by a factor of k. The x-coordinate of the vertex remains the same. The y-coordinate of the vertex changes. The x-intercepts remain the same.

Key Concepts, continued : Replacing f(x) with k f(x) and f(k x) Reflections over the x-axis: when k = –1 in k f(x) The parabola is reflected over the x-axis. The x-coordinate of the vertex remains the same. The y-coordinate of the vertex changes. The x-intercepts remain the same. When k < 0, first perform the vertical stretch or compression, and then reflect the function over the x-axis.

Key Concepts, continued Multiplying the Independent Variable by a Constant, k: f(k x) In general, multiplying the independent variable in a function by a constant will stretch or shrink the graph of f horizontally. If k > 1, the graph of f(x) will shrink or compress horizontally by a factor of (so the parabola will appear narrower). A horizontal compression squeezes the parabola toward the y-axis : Replacing f(x) with k f(x) and f(k x)

Key Concepts, continued If 0 < k < 1, the graph of f(x) will stretch horizontally by a factor of (so the parabola will appear wider). A horizontal stretch pulls the parabola and stretches it away from the y-axis. If k < 0, the graph is first horizontally stretched or compressed and then reflected over the y-axis. The y-intercept remains the same, as does the y-coordinate of the vertex : Replacing f(x) with k f(x) and f(k x)

Key Concepts, continued When a constant k is multiplied by the variable x of a function f(x), the interval of the intercepts of the function is increased or decreased depending on the value of k. The roots of the equation ax 2 + bx + c = 0 are given by the quadratic formula, Remember that in the standard form of an equation, ax 2 + bx + c, the only variable is x; a, b, and c represent constants : Replacing f(x) with k f(x) and f(k x)

Key Concepts, continued If we were to multiply x in the equation ax 2 + bx + c by a constant k, we would arrive at the following: Use the quadratic formula to find the roots of, as shown on the following slide : Replacing f(x) with k f(x) and f(k x)

Key Concepts, continued : Replacing f(x) with k f(x) and f(k x) Quadratic equation Substitute ak 2 for a and bk for b based on the equation in standard form. Simplify.

Key Concepts, continued : Replacing f(x) with k f(x) and f(k x) Horizontal compressions: when k > 1 in f(k x) The graph is compressed horizontally by a factor of The y-coordinate of the vertex remains the same. The x-coordinate of the vertex changes.

Key Concepts, continued : Replacing f(x) with k f(x) and f(k x) Horizontal stretches: when 0 < k < 1 in f(k x) The graph is stretched horizontally by a factor of The y-coordinate of the vertex remains the same. The x-coordinate of the vertex changes.

Key Concepts, continued : Replacing f(x) with k f(x) and f(k x) Reflections over the y-axis: when k = –1 in f(k x) The parabola is reflected over the y-axis. The y-coordinate of the vertex remains the same. The x-coordinate of the vertex changes. The y-intercept remains the same. When k < 0, first perform the horizontal compression or stretch, and then reflect the function over the y-axis.

Common Errors/Misconceptions thinking that multiplying the dependent variable by a constant k yields the same equation as multiplying the independent variable by the same constant k in all cases (i.e., that k f(x) is always equal to f(k x), but this is not always true) forgetting to substitute all values of x with k x when working with the transformation f(k x) forgetting to square the constant k when substituting f(k x) into ax 2 + bx + c confusing horizontal with vertical transformations and vice versa confusing stretches and compressions : Replacing f(x) with k f(x) and f(k x)

Guided Practice Example 1 Consider the function f(x) = x 2, its graph, and the constant k = 2. What is k f(x)? How are the graphs of f(x) and k f(x) different? How are they the same? : Replacing f(x) with k f(x) and f(k x)

Guided Practice: Example 1, continued 1.Substitute the value of k into the function. If f(x) = x 2 and k = 2, then k f(x) = 2 f(x) = 2x : Replacing f(x) with k f(x) and f(k x)

Guided Practice: Example 1, continued 2.Use a table of values to graph the functions : Replacing f(x) with k f(x) and f(k x) xf(x)f(x)k f(x) –248 –

Guided Practice: Example 1, continued : Replacing f(x) with k f(x) and f(k x)

Guided Practice: Example 1, continued 3.Compare the graphs. Notice the position of the vertex has not changed in the transformation of f(x). Therefore, both equations have same root, x = 0. However, notice the inner graph, 2x 2, is more narrow than x 2 because the value of 2 f(x) is increasing twice as fast as the value of f(x). Since k > 1, the graph of f(x) will stretch vertically by a factor of 2. The parabola appears narrower : Replacing f(x) with k f(x) and f(k x) ✔

Guided Practice: Example 1, continued : Replacing f(x) with k f(x) and f(k x)

Guided Practice Example 2 Consider the function f(x) = x 2 – 81, its graph, and the constant k = 3. What is f(k x)? How do the vertices and the zeros of f(x) and f(k x) compare? : Replacing f(x) with k f(x) and f(k x)

Guided Practice: Example 2, continued 1.Substitute the value of k into the function. If f(x) = x 2 – 81 and k = 3, then f(k x) = f(3x) = (3x) 2 – 81 = 9x 2 – : Replacing f(x) with k f(x) and f(k x)

Guided Practice: Example 2, continued 2.Use the zeros and the vertex of f(x) to graph the function. To find the zeros of f(x), set the function equal to 0 and factor : Replacing f(x) with k f(x) and f(k x)

Guided Practice: Example 2, continued The zeros are –9 and : Replacing f(x) with k f(x) and f(k x) x 2 – 81 = 0 Set the function equal to 0. (x + 9)(x – 9) = 0 Factor using the difference of two squares. x + 9 = 0 or x – 9 = 0 Use the Zero Product Property. x = –9 or x = 9Solve for x.

Guided Practice: Example 2, continued : Replacing f(x) with k f(x) and f(k x) The vertex of f(x) is (0, –81). This can be seen as the translation of the parent function f(x) = x 2. The parent function is translated down 81 units.

Guided Practice: Example 2, continued 3.Using the zeros and the vertex of the transformed function, graph the new function on the same coordinate plane. Set the transformed function equal to 0 and factor to find the zeros : Replacing f(x) with k f(x) and f(k x)

Guided Practice: Example 2, continued The zeros are –3 and : Replacing f(x) with k f(x) and f(k x) 9x 2 – 81 = 0 Set the function equal to 0. 9(x 2 – 9) = 0 Use the GCF to factor out 9. 9(x + 3)(x – 3) = 0 Use the difference of two squares. x + 3 = 0 or x – 3 = 0 Use the Zero Product Property. x = –3 or x = 3Solve for x.

Guided Practice: Example 2, continued : Replacing f(x) with k f(x) and f(k x) The vertex is the same as the original function, (0, –81). This again can be seen as the translation of 9x 2 down 81 units.

Guided Practice: Example 2, continued 4.Compare the graphs. Notice the position of the vertex has not changed in the transformation of f(x). However, notice the inner graph, f(3x) = 9x 2 – 81, is narrower than f(x). Specifically, the roots are x = –9 and 9 for f(x) and x = –3 and 3 for f(3x). This is because the roots of the function f(k x) are times the roots of f(x) in a quadratic equation : Replacing f(x) with k f(x) and f(k x)

Guided Practice: Example 2, continued Since k > 1, the graph of f(x) will shrink horizontally by a factor of, so the parabola appears narrower : Replacing f(x) with k f(x) and f(k x) ✔

Guided Practice: Example 2, continued : Replacing f(x) with k f(x) and f(k x)