8 + 6) x 2 ) y = h(x)

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Joel Young
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. a. Horizontal shift 6 left and vertical shift up. Notice B' is ( 6, ) and B is (0, 0). b. h(x) = 0.

If x = 8, h(x) = 0.5(x + 6) + = 0.5( = 0.5( = 0.5(8) + = 0. d. Use the formula. It crosses the y-axis when x = 0. h(0) = 0.5(0 + 6) + =.

Notice B' is (5, ) and B is (0, 0). b. g(x) = 0.5(x 5) (Enter in a grapher to check.) c. Notice C' to see g(x) crosses the x-axis at 7. d.

5x CHILD y = g(x) A PARENT PARENT CHILD. a. The graph of y = a(x) is a horizontal shift of the graph of y = to the right 6 so a(x) = f(x 6).

1 . a. Horizontal shift 6 left and vertical shift up. Notice B' is ( 6, ) and B is (0, 0). b. h(x) = 0.5(x + 6) + (Enter in a grapher to check.) c. Use the graph. Notice A' to see h(x) crosses the x-axis at 8. Check with the formula. If x = 8, h(x) = 0.5(x + 6) + = 0.5( = 0.5( = 0.5(8) + = 0. d. Use the formula. It crosses the y-axis when x = 0. h(0) = 0.5(0 + 6) + =. You can also use the table ) + ) + A' y = h(x) B' C' B y 0.5x C. a. Horizontal shift 5 right and vertical shift down. Notice B' is (5, ) and B is (0, 0). b. g(x) = 0.5(x 5) (Enter in a grapher to check.) c. Notice C' to see g(x) crosses the x-axis at 7. d. Use the formula. C It crosses the y-axis when x = 0. x = 0. B C' g(0) = 0.5(0 5) = 66.5 You can also use the table. A B' A' y 0.5x CHILD y = g(x) A PARENT PARENT CHILD. a. The graph of y = a(x) is a horizontal shift of the graph of y = to the right 6 so a(x) = f(x 6). a(x) = f(x 6) b. The graph of y = b(x) is a horizontal and vertical reflection of the graph of y = so b(x) = f( x). f( x) f( x) b(x) = f( x) f( x) is a horizontal reflection of. f( x) is a vertical reflection of f( x).

c. The graph of y = c(x) is a horizontal reflection, followed by a vertical compression by a factor of followed by a vertical shift down

The graph of y = d(x) is a horizontal and vertical reflection, followed by a vertical shift up 6.

d(x) = f( x) + 6 f( x) is a vertical shift of down. f( x) f( x) f( x). Suppose the point P(, ) is a point on the graph of y = a.

2 c. The graph of y = c(x) is a horizontal reflection, followed by a vertical compression by a factor of followed by a vertical shift down units, so c(x) = f( x)., f( x) f( x) f( x) f( x) c ( x) f ( x) d. The graph of y = d(x) is a horizontal and vertical reflection, followed by a vertical shift up 6. This is the graph of b(x) shifted up 6 units. f( x) is a horizontal reflection of. f( x) is a vertical compression of f( x). d(x) = f( x) + 6 f( x) is a vertical shift of down. f( x) f( x) f( x). Suppose the point P(, ) is a point on the graph of y = a. Suppose is even: i. Report the coordinates of another point Q, which corresponds to P. (, ) ii. Plot the point Q on the grid provided. P Q b. Suppose is odd: i. Report the coordinates of another point Q, which corresponds to P. (, ) ii. Plot the point Q on the grid provided. Q P

3 5. The graph of y =.is shown. Use the graph of to write g(x) as a transformation of. Find a formula for g(x) in terms of. a. g(x) The outputs of g(x) are larger than those for so it is a vertical stretch. Compare maximum points. The graph of g(x) is a vertical stretch of the graph of by a factor of k, where k =. Thus k = and g(x) =. b The outputs of g(x) are smaller than those for so it is a vertical shrink. Compare maximum points. The graph of g(x) is a vertical compression of the graph of by a factor of k, where 80k =. You could also compare minimum points: 0k = 0. In either case, k = 0.75 and g(x) = Solve e x = 7. EXACT: x ln7. APPROXIMATE: x.85 We have e x = 7. Since the base is e, take natural logarithms of both sides. ln e x = ln 7. Use the inverse property ln e Q = Q x ln7. Check: If x.85 and e x.85 = 7., then e 7. g(x)

4 7. a. 5ln(x) = 0 8. a. EXACT: x e or APPROXIMATE: x 8.99 e We have 5ln(x) = 0 Divide both sides by 5. ln(x) = Make both sides a power of e. e ln(x) = e Use inverse property x = e Divide both sides by x e e or Check: If x 8.99 and 5ln(x) = 0, then 5ln( 8.99) 0 b. 5log x + 7 = 0 EXACT: x = 0 /5 APPROXIMATE: x.98 We have 5log x + 7 = 0 Subtract 7 from both sides. 5log x = Divide both sides by 5. log x = Make both sides a power of logx = 0 /5 Use inverse property. x = 0 /5.98 Check: If x.985 and 5log x + 7 = 0, then log(.98) u u Since this is a quadratic equation with three terms, get 0 on one side of the equation. u u 0 We have a trinomial. One way to solve this equation is to try to factor. ( + )( + ) = u u = 0 FIRST TERMS The product of both of these is u so we have u and u in the blanks. ( u + )( u + ) = u u = 0 LAST TERMS The product of both of these is so we have and in the blanks. But which goes where? The sum of the product of the inner and the product of the outer is the middle term u. One of u and u must be negative, and the sum must be u. This is only possible if u is negative and u is positive. This means u is multiplied by. ( u )( u ) = u u = 0 u + u u Thus we have (u + )(u ) = 0. Use the zero product property A B 0 A 0 or B 0 Now set each factor equal to 0 and solve. (u + ) (u ) = 0 u 0 u 0 u u

The solutions are The solution process can be enhanced by using a grapher. In the Y= Editor, enter.

This can be confirmed with a table: The zeros of y = x x are the solutions to x x = 0 b.

We have 5u = Divide both sides by 5. u = 5 Take square roots of both sides.

5 The solutions are The solution process can be enhanced by using a grapher. In the Y= Editor, enter. The zeros of the graph appear to be and. u,. This can be confirmed with a table: The zeros of y = x x are the solutions to x x = 0 b. 5u This is a quadratic equation, but since it contains only u we can divide by 5 and take square roots. We have 5u = Divide both sides by 5. u = 5 Take square roots of both sides. Remember there are two square roots. u = 5 5 Alternatively, you can get 0 on one side and factor, and use the zero product property A B 0 A 0 or B 0 5u 5u 0 (5u )(5u ) 0 5u 0 5u 0 u u 5 5 Then set each factor equal to 0.

6 A common error is to only report the positive solution and forget there is a negative square root. A quick sketch of y = 5x confirms there are two zeros. A sketch may just be done with pencil and paper using knowledge of transformations. The graph of y = 5x is a vertical shift of the graph of y = 5x down units. The graph of y = 5x is a vertical stretch of the graph of y = x by a factor of 5. c. 5u u This is a quadratic equation. Get 0 on one side of the equation. Factor. We have 5u u = 0 Factor out the greatest common factor, which here is u. u(5u ) = 0 Set each factor equal to 0 using the zero product property. u 0 5u 0 u 5 There are two solutions, namely u = 0, 5 A common error is to divide both sides of Don t do this. You will lose a solution.. 5u uby u. d. uu ( ) 0 This is a quadratic equation, but how nice! It is already in factored form. This can be solved by setting each factor equal to 0 and applying the zero product property A B 0 A 0 or B 0. We have u = 0 and u = 0. To solve u = 0 we just divide both sides by to get u = 0. To solve u = 0, add to both sides. The two solutions are u = 0,. (This can also be done just by inspection.) A common error is to multiply these out to get u 8u = 0. This is similar to part c. But unfortunately that takes you into the wrong direction. (Multiplying out is like returning clean clothes out of the dryer and putting them back in the washer. Oops! ) Since it is already factored, you do not want to reverse the factorization by distributing. 9. a. Choice B is true. Linear functions grow by a constant rate, and exponential functions grow by a constant percent rate. b. Jonesville: P = t c. Smithville: P = 5000(.0) t 0. a. The function P is exponential. P = 00(.) x. The function Q is linear. Q = x b. x. years

The equation 00(.) x = 00 + 00x is not possible to solve algebraically. Method : Using a table, enter the formulas Y= 00(.)^x and Y = 00 + x in Y= and scroll. Eventually set your step size to 0.

7 The equation 00(.) x = x is not possible to solve algebraically. Method : Using a table, enter the formulas Y= 00(.)^x and Y = 00 + x in Y= and scroll. Eventually set your step size to 0.0 y = 00(.) x y = 00+00x Method : Using a graph, set a viewing window (aided by the table) and then find the intersection point.. Use the Pythagorean Theorem. x 0 x 00 x 96 x 96 Since x is the length of a side, it is the positive square root. a. The x-coordinate of the intersection point of R and C is x =. y 0 b. The slope of P(x) is. x The vertical intercept is (0, 0). So P(x) = x and Choice C is correct. y 70 The slope of R(x) is. x The vertical intercept is (0, 0). So R(x) = x y 80 The slope of C(x) is 8. x The vertical intercept is (0, 0). So C(x) = 8x Another method: Since P = R C, you can also find C once you know P and R. P + C = R so x 0 + C = x Add 0 to both sides and subtract x from both sides. Thus C = 8x + 0.

(a) Write down the value of q and of r. (2) Write down the equation of the axis of symmetry. (1) (c) Find the value of p. (3) (Total 6 marks)

1. Let f(x) = p(x q)(x r). Part of the graph of f is shown below. The graph passes through the points ( 2, 0), (0, 4) and (4, 0). (a) Write down the value of q and of r. (b) Write down the equation of