Quadratic Function (second degree)f(x) = a (x -h) 2 + k (h, k) (h, k) Second Degree or Quadratic Function vertex at (h, k) a > 0, curve opens upwards, has a minimum. a < 0, curve opens downward, has a maximum. axis of symmetry: x = h Domain: R Range: if a > 0 : [k, ∞ ] Range: if a < 0 : [-∞ , k ] .Quadratic Function Rule Templates f (x) = a(x – x1)( x – x2) x1 and x2 are zeros. x = h is ½ way between. k = f (h) f (x)= ax2 + bx + c c is the y-intercept factor trinomial for zeros use h = – b/2a and k = f (h) to change to standard form. f (x) = a( x – h ) 2 + k given (h, k), point (x1, y1) plug in (x1, y1) to find a. standard form general form zeros form Example 1 Find the rule of correspondence in all 3 forms for a parabola with vertex at (1, –9) and y-intercept of – 8. Solution: Since we know the vertex, we'll start with the standard or function form. f(x) = a( x –h ) 2 + k f(x) = a( x –1 ) 2 u –9 Now we plug in the coordinates of the y-intercept (0, –8) to get a –8 = a( 0 –1 ) 2 –9 u a = 1 So f(x) = ( x – 1 )² – 9 To get the general form, expand the brackets and collect terms. f(x) = ( x –1 ) 2 –9 f(x) = x 2 u –2x –8 To get the zeros form, factor the trinomial. f(x) = (x –4)(x + 2) so the zeros are x = 4 and x = –2 . Quadratic: finding the rule, word problemsExample 2 The cross section of a riverbed is a parabola, 60 meters wide with a maximum depth of 6 meters. At what distance from the shores must we place 2 buoys to mark where the depth is 4 meters? (Hint: use the y-axis as the axis of symmetry.) .Since the river is 60 m wide, each shore is 30 m from the vertex or deepest point. This puts the zeros at (–30, 0) and (30, 0) Since the lowest point (maximum depth) is 6 m below the surface, the vertex is at (0, – 6). Now we find the rule of correspondence. 30 -30 (0, -6) 4 4 We need the x-values that make f(x) = – 4 The zeros form of this rule is: Substituting (0, – 6) we get – 6 = a(0 + 30)(0 – 30) f (x) = (x + 30)(x – 30) 1 150 f (x) = a(x + 30)(x – 30) Now set f (x) = – 4 and solve for x. a = 1 150 x = ±10 3 The buoys should be placed 30 – 1 0 3 meters from the shore. 30 10 3 .Hint: With any type of function, once we have the rule of correspondence, the only thing left to figure out is x-values when we're given y-values, or vice versa. In the previous example that's what we had. We wanted the x's that go with y = – 4. Sometimes, we get x-values and we have to find the corresponding y's. reminder: f (x) is just an alias for y, so we substitute the given values for x in the rule and calculate the resulting y-value. . Quadratic: finding the rule, word problemsPractice 1) Find the rule of correspondence for a parabola: a) with vertex at (1, 1) and y-intercept of 5. b) with zeros at – 1 and 3, passing through P(7, – 16). (click for solution) .2) Find the zeros for these parabolas. a) f (x) = 2x 2 + 5x – 3 b) f (x) = – (x + 1) 2 + 4 c) f (x) = – 5x 2 + 4x + 1 d) f (x) = 2(x 2 –8x + 17) (click for solution) .3) Word problems with quadratics: (make a diagram where needed!) a) A room measures 5 m by 6 m. A rectangular rug leaves an exposed border of equal width on all sides, and covers of the floor area. What are the dimensions of the rug? (click for solution) 23 .b) El Cheapo High School awarded a total of $1440 in prize money to students at the end of the year. If there had been 10 less winners, each prize would be $2 greater. How many students received prize money? (click for solution) .c) The initial price of a share of Miscumbooberated Products was $8. It fell for 25 months and reached a minimum of $3. Then it began to rise. If the price of a share follows a parabolic curve, how many months will it take for it to be $6 for the second time? (diagram!!) (click for solution) .d) Raymond rents cabins during ice fishing season. In 1997 he charged $75/day and had an average of 50 customers per day. In 1998, Raymond decided to lower the daily rental rate by x dollars and he expects that the number of daily customers will rise by twice the value of x. What is the maximum daily income that Raymond can earn in 1998 with the new conditions? (click for solution). Quadratic: finding the rule, word problemse) The cross-section of a sculpture is depicted in the diagram. B C A ? The equation of the parabola is y = – x ² + 36x – 284. The equation of the line BC is y = 3x – 24. What is the measure of segment AB? (click for solution) Solutions 1) a) vertex (1, 1), y-intercept 5. f (x) = a( x – h ) ² + k u f (x) = a( x – 1 )² + 1 Now we plug in the coordinates of the y-intercept (0, 5) to get 5 = a( 0 – 1 )² + 1 u a = 4 So f (x) = 4( x – 1 )² + 1 .b) with zeros at – 1 and 3, passing through P(7, – 16). f (x) = a(x – x1 )( x – x2) u f (x) = a(x + 1)( x – 3) Now we plug in the coordinates of P(7, – 16) –16 = a(7 + 1)( 7 –3) u a = –½ f (x) = –½ (x+ 1)( x – 3) .2) a) f (x) = 2x² + 5x – 3 2x² + 5x – 3 = 0 u (2x – 1)( x + 3) = 0 x = ½, or x = – 3 . Quadratic: finding the rule, word problemsb) f (x) = – (x + 1) 2 + 4 (x + 1) 2 = 4 u x + 1 = ± 2 x = – 3 or x = 1 .c) f (x) = – 5x 2 + 4x + 1 0 = 1 + 4x –5x 2 u (1 + 5x)(1 – x) = 0 x = –1 /5 or x = 1 .d) f (x) = 2(x 2 – 8x + 17) since b 2 – 4ac = – 4 < 0, there are no zeros. the parabola has a minimum value > 0. .3/a) 5 m 6 m x x 6 -2x 5 -2x The rug is 5 m by 4 m. (1 -2x)(5 -x) = 0 5 -11x + 2x = 0 2 30 -22x + 4x 2 = 20 (6 -2x)(5 -2x) = 20 10 -22x + 4x 2 = 0 x = ½ m since x = 5 would make a very small rug. The rug's area is 2/3(30) = 20 square meters. .. Quadratic: finding the rule, word problemsb) Let x = # of students who won a prize therefore 1440 x = the amount of each prize. with 10 less winners, each prize would be 1440 x − 10 so 1440 x + 2 = 1440 x − 10 Multiply through by the lcd x(x – 10) to get 1440(x – 10) + 2x(x – 10) = 1440x x 2 – 10x – 7200 = 0 u (x – 90)( x + 80) = 0 Ninety (90) students shared the prize money. .c) We know that (0, 8) is the y-intercept and (25, 3) is the vertex. Set up the standard form template: f (x) = a( x – h ) 2 + k u f (x) = a( x – 25 ) 2 + 3 Now substitute (0, 8) for x and y to get: 8 = a( 0 –25 ) 2 + 3 So a = 1/125 which makes f (x) = 1/125 ( x – 25 ) 2 + 3 We need the larger of 2 x-values that make f (x) or y = 6 so set y = 6. 6 = 1/125 ( x – 25 ) 2 + 3 u 375 = ( x – 25 ) 2 so x = 44.36 or 45 months d) 75 50/day (75 – x) (50 + 2x) rent ($) # of renters rent ($) # of renters 1977 1978 The income function is I(x) = (75 – x)(50 + 2x). We want h, the x-value of the vertex. We can see that the zeros are 75 and – 25, and we know h is half way between –25 25 75 (25, 5000) # of customers/day $/day so h = 25 = x. This means the max income happens when x = 25 and f (x) = $5000 Raymond's maximum daily income will be $5000 per day if he lowers the rent by $25. Quadratic: finding the rule, word problemse) What we need here is the y-value of point B, the 2nd point of intersection. B C A ? Let's find the points of intersection for the parabola and the line: y = – x ² + 36x – 284 and y = 3x – 24 u 3x – 24= – x² + 36x – 284 or x² – 33x + 260 = 0 This trinomial factors as: (x – 20)(x – 13) = 0. Since we want the larger one, x = 20. Substituting into y = 3x – 24, we get y = 36. AB = 36. Functions MathRoom Index (all content of the MathRoom Lessons © Tammy the Tutor; 2004 -). Quadratic: finding the rule, word problems