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The powers of General Form

The powers of General Form. Probe. Below are 5 different ways of representing a quadratic relationship. Four of them represent the SAME quadratic relationship. a) Find and correct the odd-one-out. b) Name the 5 forms. Probe.

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The powers of General Form

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  1. The powers of General Form

  2. Probe Below are 5 different ways of representing a quadratic relationship. Four of them represent the SAME quadratic relationship. a) Find and correct the odd-one-out. b) Name the 5 forms

  3. Probe Below are 5 different ways of representing a quadratic relationship. Four of them represent the SAME quadratic relationship. a) Find and correct the odd-one-out. b) Name the 5 forms mapping rule equation in general form equation in transformational form graphical (parabola) table of values

  4. General Form toTransformational Form Let’s look at an example.We will take the quadratic function given in general form:y = 2x2 + 12x – 4and turn it into transformational form: 1 We know that in transformational form the coefficient of ‘x’ is 1 STEP 1:Divide every term by a.

  5. General Form toTransformational Form Let’s look at an example.We will take the quadratic function given in general form:y = 2x2 + 12x – 4and turn it into transformational form: STEP 2:Move the non-x term over

  6. Completing the Square We’re getting close, but the right-hand side of the equation needs to be a perfect square: (x – HT)2 So far we have: x2 + 6x This looks like: x2 x x x x x x

  7. Completing the Square x2 x x x x x x The equivalent to making x2 + 6x a perfect square is to arrange these tiles into a square shape... x x x x2 x x x Oops…

  8. Completing the Square x2 x x x x x x Try again… Closer…. We just need tocomplete this square… x2 x x x x x We were missingnine 1×1 squares. x

  9. Completing the Square ---------- x + 3 ----------- This looks promising:We have a square with length and width dimensions of (x + 3). We just had to add 9, or 32. ---------- x + 3 ----------- x2 x x x But how can we justify this “just add 9”? Add it to both sides of the equation! But why 32? It’s half of the coefficient of x, squared. x x x

  10. General Form toTransformational Form Back to our example.We will take the quadratic function given in general form:y = 2x2 + 12x – 4and turn it into transformational form: STEP 3:Complete the square(take half the coefficient of x, square it and add it to both sides)

  11. General Form toTransformational Form Back to our example.We will take the quadratic function given in general form:y = 2x2 + 12x – 4and turn it into transformational form: STEP 4:Factor out the 1/a term on the left hand side

  12. General Form toTransformational Form Back to our example.We will take the quadratic function given in general form:y = 2x2 + 12x – 4and turn it into transformational form: Therefore, the transformational form of the quadratic functiony = 2x2 + 12x – 4 is: Now we know that its VS = 2, HT = –3, and VT = –22 From this we know its vertex, range, axis of symmetry, etc vertex is (–3, –22) range is{y≥–22, y R} axis of symmetry isx = –3

  13. Practice 1. Complete the square to find the vertex of the functiony = 2x2 – 8x + 2. Sketch its graph. 2. Complete the square to find the range of the functiony = –x2 – 5x +1? 3. Put the equation y = 0.5x2 – 3x – 1 into transformational form by completing the square.

  14. Practice: Solutions Vertex: (2, –6) Complete the square to find the vertex of the functiony = 2x2 – 8x + 2. Sketch its graph.

  15. Practice: Solutions range is{y≤ 7.25, y R} 2. Complete the square to find the range of the functiony = –x2 – 5x +1?

  16. Practice: Solutions 3. Put the equation y = 0.5x2 – 3x – 1 into transformational form by completing the square.

  17. A shortcut…eventually STEP 1:Divide every term by a. STEP 2:Move the non-x term over STEP 3:Complete the square(take half the coefficient of x, square it and add it to both sides) STEP 4:Factor out the 1/a term on the left hand side But completing the square is a lot of work…. Let’s find a shortcut to finding vertex ect. by completing the square just ONE more time, but with the general equation

  18. A shortcut…eventually STEP 1:Divide every term by a. STEP 2:Move the non-x term over STEP 3:Complete the square(take half the coefficient of x, square it and add it to both sides) STEP 4:Factor out the 1/a term on the left hand side

  19. Here’s the shortcut We now see that ANY general quadratic equationy = ax2 + bx + c can be written as This may look complicated, but is VERYhelpful… from this we can see that the HT for any general quadratic is For example: What is the axis of symmetry of the function y = –2x2 + 4x + 6? To complete the square on this takes time. But…

  20. Here’s the shortcut We now see that ANY general quadratic equationy = ax2 + bx + c can be written as This may look complicated, but is VERYhelpful… from this we can see that the VT for any general quadratic is For example: What is the vertex of the function y = –2x2 + 4x + 6? To complete the square on this takes time. But… vertex (1, 8)

  21. Here’s the shortcut We now see that ANY general quadratic equationy = ax2 + bx + c can be written as This may look complicated, but is VERYhelpful… from this we can see that the VS for any general quadratic is a. Since the VS = –2 we can graph from the vertex (1, 8): Over 1 down 2 Over 2 down 8 Over 3 down 18 For example: Graph the functiony = –2x2 + 4x + 6. To complete the square on thistakes time. But…

  22. Shortcut within a shortcut Instead of memorizing the formula for the y-coordinate of the vertex (VT): we can calculate it using the general form of the equation and thex-coordinate of the vertex (HT): For example: What is the maximum value of the function y = –2x2 + 4x + 6? Max value is y = 8.

  23. General form shortcuts: Practice For the following quadratic functions, find the vertex, sketch its parabola, give its axis of symmetry, give its range, and write in transformational form, all WITHOUT completing the square.

  24. General form shortcuts: Practice Solutions vertex (1, –2), VS = 1/4 axis of symmetry: x = 1range: {y≥ 2, y R}

  25. General form shortcuts: PracticeSolutions vertex (–4, –6), VS = 3 axis of symmetry: x = –4range: {y≥ –6, y R}

  26. Solving for x given y using general form Given a quadratic function in general form y = ax2 + bx + c and a value for y, it is not a straightforward task to find the corresponding x-value(s) because we cannot easily isolate x when it appears in 2 different forms: x and x2. Subbing in the required y-value to a quadratic function, and bringing all the terms to one side of the equation yields a quadratic equation. A quadratic equation in general form looks like this:ax2 + bx + c = 0 where a≠ 0. Solving a quadratic equation for x is also calledfinding its roots or finding its zeros.

  27. Solving for x given y using general form NOTE: the c value of the quadratic equation might be different from the c value of the quadratic function. See the following example. Example 1:Give the quadratic equation obtained from the functiony = x2 – 2x – 15 when y = –12. Although the quadratic function had coefficients a = 1, b = –2, and c = –15,subbing in −12 for y gives the quadratic equation with coefficientsa = 1, b = –2, and c = –3 Solution: −12 = x2 – 2x – 15 0 = x2 – 2x – 3

  28. Solving for x given y using general form NOTE: the c value of the quadratic equation is the SAME as the c value of the quadratic function when we are using y = 0. See the following example. Example 2:Give the quadratic equation obtained from the functiony = x2 – 2x – 15 when y = 0. Both the quadratic function and the resulting quadratic equation have coefficients a = 1, b = –2, and c = –15. Solution: 0 = x2 – 2x – 15 While this is only true when we use y = 0, this is a very common case, since we are often interested in the x-intercepts of a quadratic function.

  29. Solving for x-intercepts using general form Example 3:Find the x-intercepts of the quadratic function y = x2 – 2x – 15 Solution: 0 = x2 – 2x – 15 …but now what? Soon we will learn how to do this directly from general form…. Last class we learned how to solve this if the function had been given in transformational form… So lets try that!

  30. Solving for x-intercepts using general form STEP 1: Write in transformational form by completing the square STEP 2: Set f(x) = y = 0 STEP 3: Simplify left-hand side STEP 4: Take the squarrootof both sides STEP 5: Isolate x in both equations Therefore the x-intercepts of the parabola are (5, 0) and (–3, 0), and the roots of the function are 5, and –3.

  31. Solving for x-intercepts using general formPractice Find the roots of the following quadratics: Not worth it! Wait for the shortcut!

  32. A shortcut…eventually STEP 1: Write in transformational form by completing the square STEP 2: Set f(x) = y = 0 STEP 3: Simplify left-hand side STEP 4: Take the square-root of both sides STEP 5: Isolate x in both equations But this is a lot of work…. Let’s find a shortcut to finding x-intercepts by doing this just ONE more time, but with the general equation.

  33. A shortcut…eventually STEP 1: Write in transformational form by completing the square STEP 2: Set f(x) = y = 0 STEP 3:Simplify left-hand side

  34. A shortcut…eventually STEP 4: Take the square-root of both sides STEP 5: Isolate x in both equations This is the shortcut, the Quadratic Root Formula!

  35. Solving for x-intercepts using general formThe short-cut Back to our example:Find the x-intercepts of the graph of the functionf(x) = x2 – 2x – 15 Solution: Instead of those 5 steps, let’s apply the Quadratic Root Formula.

  36. Solving for x given y using general form The quadratic root formula can be used to find x-values other than the x-intercepts For the function f(x) = x2 – 2x – 15, find the values of x when y = −12: Solution: –12 = x2 – 2x – 15 0 = x2 – 2x – 3. Now use the quadratic root formua witha = 1, b = –2 and c = –3

  37. Solving for x-intercepts using general formPractice Find the roots of the following quadratics: Use the quadratic root formula!

  38. Finding roots from General Form: Practice Solutions

  39. Finding roots from General Form: Practice Solutions

  40. Finding roots from General Form: Practice Solutions Find the x-intercepts of the following quadratic functions: Answers:

  41. Yet another method of finding roots: Factoring As slick as the quadratic root formula is for finding roots, sometimes factoring is even faster! The zero property: If (s)(t) = 0 then s = 0, or t = 0, or both.(This only works when the product is 0.) So if we can get the quadratic function in the form:y = (x – r1)(x – r2), (called factored form)then when y = 0 we know that: (x – r1) = 0, or (x – r2) = 0, or both. In other words, x = r1 or x = r2 or both.

  42. Factoring to find the roots Ex. Find the x-intercepts of the graph of the functionf(x) = x2 – 2x – 15 Answer when the numbers are multiplied Answer when the numbers are added We need to factor x2 – 2x – 15.We need 2 numbers whose product is –15 and whose sum is –2 (There is never more than one pair of numbers that work!) -5 3 ___ ×___ = –15 ___ + ___ = – 2 -5 So when looking for roots, 0 = (x – 5)(x + 3) and x = 5 or x = –3 3 So the factored form off(x) = x2 – 2x – 15 isf(x) = (x – 5)(x + 3) So the x-intercepts of the graph are (5, 0) and (–3, 0)

  43. Factoring When factoring a quadratic equation where a = 1,find two numbers that multiply to give c and add to give b. By the way: (0) = x2 + 10x – 24 24 = x2 + 10x 24 + 25 = x2 + 10x + 25 49 = (x + 5)2 ±7 = x + 5 x = –5 + 7 or x = –5 – 7 x = 2 or x = –12 Ex. Find the roots ofy = x2 + 10x – 24by factoring. 12 -2 ___ x ___ = -24 ___ + ____ = 10 -2 12 or

  44. Practice a) 0 = x2 + 2x +1 c) 0 = x2 + 2x – 24 b) 0 = x2 + 5x +4 d) 0 = x2 – 25 2. Find the x- and y- intercepts of the following quadratics. a) f(x) = 2x2 +3x + 1 c) f(x) = x2 – 5x -14 b) f(x) = 3x2 – 7x + 2 d) y = 2(x – 4)2 – 32 Answers 1a. x = –1 b. x = –4 or –1 c. x = –6 or 4 d. x = –5 and 5 2a. (–1, 0) (–0.5, 0) (0, 1) b. (1/3, 0) (2, 0) (0, 2) c. (–2, 0) (7, 0) (0, –14) d. (0, 0) (8, 0) 1. Solve the following equations by factoring.

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