Modeling with Quadratic Functions Depending on what information we are given will determine the form that is the most convenient to use. Graphing 2 y=x Standard Vertex y = ax2+bx+c y= a (x - h)2+k axis: x b 2a vertex ( x , y ) axis: x = h vertex ( h , k ) C is y-intercept a>0 U shaped, Minimum a>0 U shaped, Minimum a<0 ∩ shaped, Maximum a<0 ∩ shaped, Maximum 1. Write a quadratic function for the parabola shown. Use vertex form because the vertex is given. y = a(x – h)2 + k Vertex form y = a(x – 1)2 – 2 Substitute 1 for h and –2 for k. Use the other given point, (3, 2), to find a. Substitute 3 for x and 2 for y. 2 = a(3 – 1)2 – 2 Simplify coefficient of a. 2 = 4a – 2 Solve for a. 1=a A quadratic function for the parabola is y = 1(x – 1)2 – 2. 2. Write a quadratic function whose graph has the given characteristics. vertex: (4, –5) passes through: (2, –1) Use vertex form because the vertex is given. Vertex form y = a(x – h)2 + k y = a(x – 4)2 – 5 Substitute 4 for h and –5 for k. Use the other given point, (2,–1), to find a. Substitute 2 for x and –1 for y. –1 = a(2 – 4)2 – 5 Simplify coefficient of x. –1 = 4a – 5 Solve for a. 1=a ANSWER A quadratic function for the parabola is y = 1(x – 4)2 – 5. 3. vertex: (–3, 1) passes through: (0, –8) Use vertex form because the vertex is given. y = a(x – h)2 + k Vertex form y = a(x + 3)2 + 1 Substitute –3 for h and 1 for k. Use the other given point, (0,–8), to find a. Substitute 2 for x and –8 for y. –8 = a(0 + 3)2 + 1 Simplify coefficient of x. –8 = 9a + 1 Solve for a. –1 = a ANSWER A quadratic function for the parabola is y = 1(x + 3)2 + 1. Steps for solving in 3 variables 1. Using the 1st 2 equations, cancel one of the variables. 2. Using the last 2 equations, cancel the same variable from step 1. 3. Use the results of steps 1 & 2 to solve for the 2 remaining variables. 4. Plug the results from step 3 into one of the original 3 equations and solve for the 3rd remaining variable. 5. Write the quadratic equation in standard form. 4. Write a quadratic function in standard form for the parabola that passes through the points (–1, –3), (0, – 4), and (2, 6). STEP 1 Substitute the coordinates of each point into y = ax2 + bx + c to obtain the system of three linear equations shown below. –3 = a(–1)2 + b(–1) + c Substitute –1 for x and -3 for y. –3 = a – b + c Equation 1 –4 = a(0)2 + b(0) + c –4=c 6 = a(2)2 + b(2) + c Substitute 0 for x and – 4 for y. 6 = 4a + 2b + c Equation 2 Substitute 2 for x and 6 for y. Equation 3 STEP 2 Rewrite the system of three equations in Step 1 as a system of two equations by substituting – 4 for c in Equations 1 and 3. a–b+c=–3 a–b–4=–3 a–b=1 4a + 2b + c = 6 4a + 2b - 4 = 6 4a + 2b = 10 Equation 1 Substitute – 4 for c. Revised Equation 1 Equation 3 Substitute – 4 for c. Revised Equation 3 STEP 3 Solve the system consisting of revised Equations 1 and 3. Use the elimination method. a–b=1 4a + 2b = 10 2a – 2b = 2 4a + 2b = 10 6a = 12 a=2 So 2 – b = 1, which means b = 1. The solution is a = 2, b = 1, and c = – 4. ANSWER A quadratic function for the parabola is y = 2x2 + x – 4. 5. Write a quadratic function in standard form for the parabola that passes through the given points. (–1, 5), (0, –1), (2, 11) STEP 1 Substitute the coordinates of each point into y = ax2 + bx + c to obtain the system of three linear equations shown below. 5 = a(–1)2 + b(–1) + c Substitute –1 for x and 5 for y. 5=a–b+c Equation 1 –1 = a(0)2 + b(0) + c –1=c Substitute 0 for x and – 1 for y. 11 = a(2)2 + b(2) + c Substitute 2 for x and 11 for y. 11 = 4a + 2b + c Equation 3 Equation 2 STEP 2 Rewrite the system of three equations in Step 1 as a system of two equations by substituting – 1 for c in Equations 1 and 3. a–b+c=5 a–b–1=5 a–b=6 4a + 2b + c = 11 4a + 2b – 1 = 11 4a + 2b = 12 Equation 1 Substitute – 1 for c. Revised Equation 1 Equation 3 Substitute – 1 for c. Revised Equation 3 STEP 3 Solve the system consisting of revised Equations 1 and 3. Use the elimination method. a–b=6 4a + 2b = 10 2a – 2b = 12 4a + 2b = 12 6a = 24 a= 4 So, 4 – b = 6, which means b = – 2 ANSWER A quadratic function for the parabola is y = 4x2 – 2x – 1 6. (1, 0), (2, -3), (3, -10) STEP 1 Substitute the coordinates of each point into y = ax2 + bx + c to obtain the system of three linear equations shown below. 0 = a(1)2 + b(1) + c 0= 1a + 1b + c Substitute 1 for x and 0 for y. Equation 1 -3 = a(2)2 + b(2) + c -3 = 4a +2b + c Substitute 2 for x and- 3 for y. -10 = a(3)2 + b(3) + c Substitute 3 for x and -10 for y. -10 = 9a + 3b + c Equation 3 Equation 2 STEP 2 Rewrite the system of three equations in Step 1 as a system of two equations. Use Elimination to simplify one set of equations then use substitution & elimination to find the PART 1 Equation 1 Equation 2 0 = 1a + 1b + c -1 0 = -1a - 1b - c -3 = 4a +2b + c -3 = 3a +1b PART 1 -1 3 = -4a -2b - c -10 = 9a + 3b + c -7 = 5a + 1b PART 2 -3 = 4a +2b + c PART 2 Equation 2 -3 = 4a +2b + c Equation 3 -10 = 9a + 3b + c PART 1 PART 2 -3 = 3a +1b -7 = 5a + 1b -1 3 = -3a - 1b -7 = 5a + 1b -4 = 2a -2 = a STEP 3 Solve for the remaining variables by substitution. Original equations 0 = 1a + 1b + c -3 = 4a +2b + c -10 = 9a + 3b + c -3 = 3a +1b -3 = 3a +1b PART 1 -7 = 5a + 1b PART 2 -2 = a -3 = 3(-2) +1b -3 = -6 +1b 3=b 0 = 1a + 1b + c 0 = 1(-2) + 1(3) + c 0 = -2 + 3 + c 0=1+c -1 = c Substitute -2 = a 3=b -1 = c y = ax2 + bx + c y= 2 -2x + 3x -1 FINAL ANSWER!! QUIZ Wednesday, February 15, 2012 One question each similar to the following examples (in this power point): Examples: 2, 5, & 6 GOOD LUCK!!

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