5 8 Modeling with Quadratic Functions Modeling with

5. 8 Modeling with Quadratic Functions

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 = ax 2+bx+c y= a (x - h)2+k axis: x = h vertex ( x , y ) vertex ( h , k ) C is y-intercept a>0 U shaped, Minimum 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 = 4 a – 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 = 4 a – 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 = 9 a + 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 1 st 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 3 rd 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 = ax 2 + 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 = 4 a + 2 b + 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 4 a + 2 b + c = 6 4 a + 2 b - 4 = 6 4 a + 2 b = 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 4 a + 2 b = 10 2 a – 2 b = 2 4 a + 2 b = 10 6 a = 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 = 2 x 2 + 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 = ax 2 + 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 = 4 a + 2 b + 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 4 a + 2 b + c = 11 4 a + 2 b – 1 = 11 4 a + 2 b = 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 4 a + 2 b = 10 2 a – 2 b = 12 4 a + 2 b = 12 6 a = 24 a= 4 So, 4 – b = 6, which means b = – 2 ANSWER A quadratic function for the parabola is y = 4 x 2 – 2 x – 1

6. (1, 0), (2, -3), (3, -10) STEP 1 Substitute the coordinates of each point into y = ax 2 + bx + c to obtain the system of three linear equations shown below. 0 = a(1)2 + b(1) + c 0= 1 a + 1 b + c Substitute 1 for x and 0 for y. Equation 1 -3 = a(2)2 + b(2) + c -3 = 4 a +2 b + 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 = 9 a + 3 b + 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 2 0 = 1 a + 1 b + c -1 -3 = 4 a +2 b + c PART 2 Equation 2 -3 = 4 a +2 b + c Equation 3 -10 = 9 a + 3 b + c PART 1 PART 2 -3 = 3 a +1 b -7 = 5 a + 1 b -1 -1 0 = -1 a - 1 b - c -3 = 4 a +2 b + c -3 = 3 a +1 b PART 1 3 = -4 a -2 b - c -10 = 9 a + 3 b + c -7 = 5 a + 1 b PART 2 3 = -3 a - 1 b -7 = 5 a + 1 b -4 = 2 a -2 = a

STEP 3 Solve for the remaining variables by substitution. Original equations 0 = 1 a + 1 b + c -3 = 4 a +2 b + c -10 = 9 a + 3 b + c -3 = 3 a +1 b PART 1 -7 = 5 a + 1 b PART 2 -2 = a -3 = 3(-2) +1 b -3 = -6 +1 b 3=b 0 = 1 a + 1 b + c 0 = 1(-2) + 1(3) + c 0 = -2 + 3 + c 0=1+c -1 = c

Substitute -2 = a 3=b -1 = c y = ax 2 + bx + c y= 2 -2 x + 3 x -1 FINAL ANSWER!!