# modeling with quadratic functions

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Modeling with Quadratic Functions. Depending on what information we are given will determine the form that is the most convenient to use. Graphing y=x 2. 1. Write a quadratic function for the parabola shown. Use vertex form because the vertex is given. y = a ( x – h ) 2 + k. - PowerPoint PPT PresentationTRANSCRIPT

Modeling with Quadratic FunctionsDepending on what information we are given will determine the form that is the most convenient to use.

Graphing y=x2

StandardVertexy = ax2+bx+cy= a (x - h)2+kaxis: axis: x = hvertex ( x , y )vertex ( h , k )C is y-intercepta>0 U shaped, Minimuma>0 U shaped, Minimuma

1. Write a quadratic function for the parabola shown.Use vertex form because the vertex is given.y = a(x h)2 + kVertex formy = a(x 1)2 2Substitute 1 for h and 2 for k.Use the other given point, (3, 2), to find a.2 = a(3 1)2 2Substitute 3 for x and 2 for y.2 = 4a 2Simplify coefficient of a.1 = aSolve 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.Use vertex form because the vertex is given.y = a(x h)2 + kVertex formy = a(x 4)2 5Substitute 4 for h and 5 for k.Use the other given point, (2,1), to find a.1 = a(2 4)2 5Substitute 2 for x and 1 for y.1 = 4a 5Simplify coefficient of x. 1 = aSolve for a.

Use vertex form because the vertex is given.y = a(x h)2 + kVertex formy = a(x + 3)2 + 1Substitute 3 for h and 1 for k.Use the other given point, (0,8), to find a.8 = a(0 + 3)2 + 1Substitute 2 for x and 8 for y.8 = 9a + 1Simplify coefficient of x. 1 = aSolve for a.

Steps for solving in 3 variablesUsing 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 1Substitute 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) + cSubstitute 1 for x and -3 for y.3 = a b + cEquation 14 = a(0)2 + b(0) + cSubstitute 0 for x and 4 for y. 4 = cEquation 26 = a(2)2 + b(2) + cSubstitute 2 for x and 6 for y.6 = 4a + 2b + cEquation 3

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.STEP 2a b + c = 3Equation 1a b 4 = 3Substitute 4 for c.a b = 1Revised Equation 14a + 2b + c = 6Equation 34a + 2b - 4 = 6Substitute 4 for c.4a + 2b = 10Revised Equation 3

a b = 12a 2b = 24a + 2b = 104a + 2b = 10a = 2So 2 b = 1, which means b = 1.The solution is a = 2, b = 1, and c = 4.STEP 3Solve the system consisting of revised Equations 1 and 3. Use the elimination method.

5. Write a quadratic function in standard form for the parabola that passes through the given points.(1, 5), (0, 1), (2, 11)STEP 1Substitute 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) + cSubstitute 1 for x and 5 for y.5 = a b + cEquation 11 = a(0)2 + b(0) + cSubstitute 0 for x and 1 for y. 1 = cEquation 211 = a(2)2 + b(2) + cSubstitute 2 for x and 11 for y.11 = 4a + 2b + cEquation 3

a b + c = 5Equation 1a b 1 = 5Substitute 1 for c.a b = 6Revised Equation 14a + 2b + c = 11Equation 34a + 2b 1 = 11Substitute 1 for c.4a + 2b = 12Revised Equation 3Rewrite the system of three equations in Step 1 as a system of two equations by substituting 1 for c in Equations 1 and 3.STEP 2

a b = 62a 2b = 124a + 2b = 104a + 2b = 12a = 4So, 4 b = 6, which means b = 2STEP 3Solve the system consisting of revised Equations 1 and 3. Use the elimination method.

6. (1, 0), (2, -3), (3, -10)STEP 1Substitute 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) + cSubstitute 1 for x and 0 for y.0= 1a + 1b + cEquation 1-3 = a(2)2 + b(2) + cSubstitute 2 for x and- 3 for y. -3 = 4a +2b + cEquation 2-10 = a(3)2 + b(3) + cSubstitute 3 for x and -10 for y.-10 = 9a + 3b + cEquation 3

Equation 1Equation 2Equation 2Equation 3Rewrite 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 STEP 20 = 1a + 1b + c -3 = 4a +2b + c-10 = 9a + 3b + c-10 = -1a - 1b - c -3 = 4a +2b + c -3 = 3a +1bPART 1PART 1 -3 = 4a +2b + cPART 2-13 = -4a -2b - c-10 = 9a + 3b + c-7 = 5a + 1bPART 2PART 1PART 2 -3 = 3a +1b-7 = 5a + 1b-1 3 = -3a - 1b-7 = 5a + 1b-4 = 2a -2 = a

STEP 3Solve for the remaining variables by substitution.0 = 1a + 1b + c-10 = 9a + 3b + c -3 = 4a +2b + cOriginal equations-2 = a -7 = 5a + 1b -3 = 3a +1bPART 1PART 2 -3 = 3a +1b -3 = 3(-2) +1b -3 = -6 +1b 3 = b0 = 1a + 1b + c0 = 1(-2) + 1(3) + c0 = -2 + 3 + c0 = 1 + c-1 = c

-2 = a 3 = b-1 = cy = ax2 + bx + c y = -2x2 + 3x -1SubstituteFINAL ANSWER!!

QUIZ Wednesday, February 15, 2012One question each similar to the following examples (in this power point):Examples: 2, 5, & 6GOOD LUCK!!

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