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 y=x2

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 forhand –2 for k. Use the other given point, (3, 2), to find a. 2= a(3– 1)2 – 2 Substitute 3 for xand 2 for y. 2 = 4a – 2 Simplify coefficient of a. 1 = a Solve for 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) ANSWER A quadratic function for the parabola is y = 1(x – 4)2 – 5. Use vertex form because the vertex is given. y = a(x – h)2 + k Vertex form y = a(x – 4)2 – 5 Substitute 4 forhand –5 for k. Use the other given point, (2,–1), to find a. –1= a(2– 4)2 – 5 Substitute 2 for xand –1 for y. –1 = 4a – 5 Simplify coefficient of x. 1 = a Solve for a.

3. vertex: (–3, 1) passes through: (0, –8) ANSWER A quadratic function for the parabola is y = 1(x + 3)2 + 1. Use vertex form because the vertex is given. y = a(x – h)2 + k Vertex form y = a(x + 3)2 + 1 Substitute –3 forhand 1for k. Use the other given point, (0,–8), to find a. –8= a(0+3)2 + 1 Substitute 2 for xand –8 for y. –8 = 9a + 1 Simplify coefficient of x. –1 = a Solve for a.

Steps for solving in 3 variables • Using the 1st 2 equations, cancel one of the variables. • Using the last 2 equations, cancel the same variable from step 1. • Use the results of steps 1 & 2 to solve for the 2 remaining variables. • Plug the results from step 3 into one of the original 3 equations and solve for the 3rd remaining variable. • 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 + cto obtain the system of three linear equations shown below. –3= a(–1)2 + b(–1) + c Substitute –1 for xand -3 for y. –3 = a – b + c Equation 1 –4 = a(0)2 + b(0) + c Substitute 0 for xand – 4 for y. – 4 = c Equation 2 6= a(2)2 + b(2) + c Substitute 2 for xand 6 for y. 6 = 4a + 2b + c Equation 3

STEP 2 Rewrite the system of three equations in Step 1 as a system of two equations by substituting – 4 for cin Equations 1 and 3. a – b + c= – 3 Equation 1 a – b – 4 = – 3 Substitute – 4 for c. a –b = 1 Revised Equation 1 4a + 2b + c= 6 Equation 3 4a + 2b - 4= 6 Substitute – 4 for c. 4a + 2b = 10 Revised Equation 3

6a = 12 ANSWER A quadratic function for the parabola isy = 2x2 + x – 4. STEP 3 Solve the system consisting of revised Equations 1 and 3. Use the elimination method. 2a – 2b = 2 a – b = 1 4a + 2b = 10 4a + 2b = 10 a = 2 So 2 – b = 1, which means b = 1. The solution is a = 2, b = 1, and c = – 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 + cto obtain the system of three linear equations shown below. 5= a(–1)2 + b(–1) + c Substitute –1 for xand 5 for y. 5= a – b + c Equation 1 –1= a(0)2 + b(0) + c Substitute 0 for xand – 1 for y. – 1 = c Equation 2 11= a(2)2 + b(2) + c Substitute 2 for xand 11 for y. 11 = 4a + 2b + c Equation 3

STEP 2 Rewrite the system of three equations in Step 1 as a system of two equations by substituting – 1 for cin Equations 1 and 3. a – b + c= 5 Equation 1 a – b – 1 = 5 Substitute – 1 for c. a –b = 6 Revised Equation 1 4a + 2b + c = 11 Equation 3 4a + 2b – 1 = 11 Substitute – 1 for c. 4a + 2b = 12 Revised Equation 3

Solve the system consisting of revised Equations 1 and 3. Use the elimination method. STEP 3 6a = 24 ANSWER A quadratic function for the parabola is 4x2– 2x – 1 y = a – b = 6 2a – 2b = 12 4a + 2b = 10 4a + 2b = 12 a = 4 So, 4 – b = 6, which means b = – 2

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 Substitute 1 for xand 0 for y. 0= 1a + 1b + c Equation 1 -3= a(2)2 + b(2) + c Substitute 2 for xand- 3 for y. -3 = 4a +2b + c Equation 2 -10= a(3)2 + b(3) + c Substitute 3 for x and -10 for y. -10 = 9a + 3b + c Equation 3

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 0 = 1a + 1b + c -1 Equation 1 0 = -1a - 1b - c -3 = 4a +2b + c Equation 2 -3 = 4a +2b + c -3 = 3a +1b PART 1 PART 2 3 = -4a -2b - c -3 = 4a +2b + c -1 Equation 2 -10 = 9a + 3b + c Equation 3 -10 = 9a + 3b + c -7 = 5a + 1b PART 2 -3 = 3a +1b -1 3 = -3a - 1b PART 1 -7 = 5a + 1b -7 = 5a + 1b PART 2 -4 = 2a -2 = a

STEP 3 Solve for the remaining variables by substitution. Original equations 0 = 1a + 1b + c -3 = 3a +1b PART 1 -3 = 4a +2b + c -7 = 5a + 1b PART 2 -10 = 9a + 3b + c -2 = a -3 = 3a +1b -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 = -2x2 +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!!

Modeling with Quadratic Functions