Convert quadratic functions from Standard Form to Intercept Form or Vertex Form (By Nghi H Nguyen Nov 30, 2014)

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1 Convert quadratic functions from Standard Form to Intercept Form or Vertex Form (By Nghi H Nguyen Nov 30, 2014) The Quadratic Function in Intercept Form The graph of the quadratic function in standard form f(x) = ax² + bx + c is a parabola that may intercept the x-axis at 2 points, unique point, or no point at all. This means a quadratic equation f(x) = ax² + bx + c = 0 may have 2 real roots, one double real root, or no real roots at all (complex roots). We can write f(x) in the form: y = a(x² + b/a x + c/a) (1). Recall the development of the quadratic formula: x² + bx/a + (b²/4a² - b²/4a²) + c/a = 0 (x + b/2a)² - (b² - 4ac)/4a² = 0 (x + b/2a)² - d²/4a² = 0. (Call d² = b² - 4ac) (x + b/2a + d/2a)(x + b/2a d/2a) = 0 (2) Replace this expression (2) into the equation (1), we get the quadratic function f(x) written in intercept form: f(x) = a(x x1)(x x2) (x1 and x2 are the 2 x-intercepts, or roots of f(x) = 0) f(x) = a(x + b/2a + d/2a)(x + b/2a d/2a) (3) The Quadratic Formula in Intercept Form. From the equation (3), we deduct the formula: x = -b/2a ± d/2a (4) In this formula, x being the 2 roots of the quadratic equation f(x) = 0: - The quantity ( b/2a) represents the x-coordinate of the parabola axis. - The 2 quantities (d/2a) and (-d/2a) represent the 2 distances AB and AC from the parabola axis to the two x-intercepts (real roots) of the parabola. - The quantity d can be zero, a number, or imaginary number. - If d = 0: there is a double root at x = -b/2a. - If d is a number (real or radical), there are two x-intercepts, meaning two real roots. - If d is imaginary, there are no real roots. The parabola doesn t intercept the x-axis.

2 - The unknown quantity d can be computed from the constants a, b, c by this relation (5): d² = b² 4ac (5) We can easily obtain this relation by writing that the product (x1*x2) of the 2 real roots is equal to (c/a). (-b/2a + d/2a)(-b/2a d/2a) = c/a b² d² = 4ac d² = b² 4ac If d ² = 0, there is double root at x = -b/2a If d² > 0, there are 2 real roots. If d² < 0, there are no real roots, there are complex roots. Page 2 of 5

3 To solve a quadratic equation, first find the quantity d by the relation (5), then find the 2 real roots by the formula (4). REMARK. This new quadratic formula in intercept (graphic) form is simpler and easier to remember than the classic formula since students can relate it to the x-intercepts of the parabola graph. In addition, the two quantities (d/2a) and (-d/2a) make more sense about distance than the classical quantity: (b^2 4ac). Converting a quadratic function from Standard Form to Intercept Form. To convert a function, in standard form f(x) = ax² + bx + c, into the intercept form, we can use the general intercept form (3). f(x) = a*(x + b/2a + d/2a)(x + b/2a d/2a), with d² = b² - 4ac. a. If d is real, whole number (3, 14, 76 ), or fraction (3/5, 23/47 ), f(x) can be factored and f(x) can be converted to intercept form. Examples: y = (x 7)(x + 8); y = (x + 3)(x + 23); y = (3x 5)(7x 13); y = (5x 7)(6x 5) b. If d is radical ( 5, 13, 71 ), f(x) can t be factored. However f(x) can be converted to intercept form. See Examples 10 and 11. Examples: y = (x 1/2 + 5/2)(x 1/2-5/2); y = (x - 2/3-3/3)(x 2/3 + 3/3) c. If d is imaginary (3i, 17i ), there are no x-intercepts and there is no intercept form. Example 7. Convert y = 5x² + 6x 8 into intercept form Solution. First find d² = = 196 d = 14 and d = Substitute the values of d, a, b, c into (3), we get y in intercept form. y = 5*(x + 6/ /10)(x + 6/10 14/10) y = 5*(x 4/5)(x + 2) = (5x 4)(x + 2). Example 8. Convert y = 7x^2 + 18x 25 into intercept form Solution. First find d² = = 1024 d = 32 and d = -32. y = a*(x 18/14-32/14)(x 18/ /14) Page 3 of 5

4 Intercept form: y = 7(x - 25/7)(x + 1) = (7x 25)(x + 1) Example 9. Convert y = 16x² + 34x 15 to standard form. Solution. First find d² = b² 4ac = = 2116 d = 46 and d = -46 Since d is a whole number, we can either factor the equation, or use the formula. Factoring gives the intercept form: y = (8x 3)(2x + 5) Example 10. Convert y = 9x² + 4x 1 Solution. d² = = 52 = 4*13 d = 2 13 and d = Replace values of d, a, b, c into the general expression (3). Intercept form: y = 9[x + (2 + 13)/9)(x + (2-13)/9). y = 1/9(9x )(9x ) We may check the answer by multiplying the 2 factors: y = 1/9[(9x + 2)² - 13] = 1/9( 81x² + 36x ) = 9x² + 4x - 1 Example 11. Convert y = 9x² - 12x + 1. Solution. d² = = 108 = 36*3 d = 6 3 and d = Intercept form: y = 9(x - 2/3 + 3/3)(x - 2/3-3/3) = (3x 2 + 3)(3x 2-3) We may check the answer by multiplying the 2 factors: y = (3x 2)² - 3 = 9x² - 12x = 9x² - 12x + 1 Converting a quadratic function from Standard Form to Vertex Form. We can convert a quadratic function, in standard form y = ax² + bx + c, into the general vertex form: y = a(x + p)² + q. We don t need to factor the quadratic equation because factoring is only a special case of finding the 2 real roots. The below method is generally better. Recall the formula (4) of the quadratic formula in intercept form: x = -b/2a d/2a. (4) The quantity (-b/2a) represents the x-coordinate of the parabola-axis. At the vertex, d = 0 since the horizontal distances of the vertex to the 2 related x- intercepts are zero. Consequently, the general vertex form of a quadratic function can be written as: f(x) = a(x + b/2a)² + f(-b/2a) (6) Example 12. Convert y = 5x² + 6x 8 into vertex form. Page 4 of 5

5 Solution. We have: a = 5; and (-b/2a) = -6/10 = -3/5; f(-3/5) = 5(9/25) 6(3/5) 8 = -49/5 The vertex form y = 5(x + 3/5)² - 49/5 Example 13. Convert y = 7x² + 18x 25 into vertex form. Solution. We have: a = 7; and (-b/2a) = -18/14 = -9/7 f(-9/7) = 7(81/49) 18(9/7) 25 = 81/7 162/7 175/7 = 256/7 The vertex form y = 7(x + 9/7)² - 256/7. Example 14. Convert y = 2x^2 + 9x + 10, into vertex form. Solution. (-b/2a) = -9/4 f(-9/4) = 2(81/16) 9(9//4) + 10 = 81/8 162/8 + 80/8 = -1/8. Vertex form y = 2(x + 9/4)² - 1/8 [This article was written by Nghi H. Nguyen, author of the new Transforming Method (Google, Yahoo, or Bing Search) for solving quadratic equations Nov 30, 2014] Page 5 of 5