3_2_Q2

3_2_Q2 - Student’s Name Signature Quiz 2: Section 3.2...

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Unformatted text preview: Student’s Name Signature Quiz 2: Section 3.2 Reminder of the Tmnsfomation of Power function: “93) = 510? —01)" + 01 1 Graphing Assignment The goal of this assignment is to ShOW you how to sketch f = —2(:I; — 1)5 + 2 without using calculators. Read my instructions from Step 1 to Step .5. At each step, sketch the graph based on the given’table which I have filled out. 0 Step 0: Write down a] = 1, bl = —2, c1 = 2 and n = 5 of the function f(av) = —2(:1: - 1)5 + 2. ‘X 0 Step 1: Graph f(x) = 9:” depending on whether 71 is ODD or EVEN. A Graph f = x5 (n = 5: ODD power) based on the following table: Table 1: Table of the function y = f (z) = 11:5 n n 0 Step 2: Check for reflection about the z-amis. Because bl = —2 < 0, reflect the graph of y = f = $5 Step 1) about the z—axis. The result of this step is the graph 3/ = f (x) = —x5. Graph y = f (x) = —x5 based on the following table: Table 2:‘Table of the function y = f (:13) = —a:5 n Ill a Step 3: Shift graph to left or right. Graph 3/ = f(:c) = sign(b1)(w — a1)" = —(a: — 1)5 based on the following table: Table 3: Table of the function y = f(a:) = —(z — 1)5 n n The graph of y = f(:13) = —(:1: — 1)5 is the result of SHIFTING the graph y = f(:c) = —:cS (in Step 2) to the RIGHT 1 unit. It is due to the fact that 0.1 = 1 > 0. 0 Step 4: Stretch or Compress graph vertically. Graph f(x) = b1(x — a1)" = —2(ac — 1)5 based on the following table: Table 4: Table of the function y = f(:z) = —2(z — 1)5 : an ” a Step 5: Shift graph up 01' down. Graph f($) = b1(:c — a1)" + 01 = —2(a: — 1)5 + 2 based on the following tablefz Table 5: Table of the function y = f(;v) = —2(:r: — 1)5 + 2 u an The graph of y = f(z) = —-2(z — 1)5 + 2 is the result of SHIFTING the graph y =. E f(:c) = —2($ — 1)5 (in Step 4) UP 2 units. It is due to the fact that c; = 2 > 0. 7 ._ —> These are the steps that you can do to sketch the graph of f(a:) = —2(z — 1)5 + 2 without using a calculator. 2 Important note For large lzl, the graph of flat) = —2($ — 1)5 + 2 behaves like the graph of f(:v) = —23:5. Lyl ...
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