Use a system of equations to solve each problem. See Example 8. F...

Question

Use a system of equations to solve each problem. See Example 8. Find an equation of the parabola y = ax^2 + bx + c that passes through the points (2, 3), (-1, 0), and (-2, 2).

Accepted Answer

Right. In the equation of the problem that passes through negative 67, negative 34 and negative 1 12 express the equation in the form of Y equals A X squared plus B X plus C. And make use of a system of equations in finding A B and C. Then we have four different equations as our possible answer. Now to solve this, let's create a system of equations using our points. Our first point we notice is negative 67. Let's set our Parabola equation with our point seven equals a multiplied by negative six squared plus B multiplied by negative six plus C. Give me a seven equals 36 A minus six B plus C. We do the same for our other two points. Either the three and four gives us four equals a multiplied by negative three squared plus negative three, multi by B plus C which gives us nine A minus three B plus C. Finally, we have negative 1 12. So negative 1 12 will be 12 equals of one squared A plus native one type B or C risk is A 12 equals A minus B plus C. We now have a system of three equations. 98 minus three B plus C equals four 36 A minus six B plus C equals seven and A minus B plus C equals 12. Now, we need to solve this system of equations to solve this system. We can make use of Kramer's rule. Let's first find matrix D matrix D will just be the coefficient of our current matrix. 36 negative nine negative 31 and one negative 11, which is the determinant of this matrix. If we were to solve the determinant, we would get negative 30. Now what our next determinant. So we can set this up by replacing column A with our last column. So if that works, let's say we want D A, we replace our first column with our last three numbers in each of our equations. 74 12. It will keep the other numbers the same negative six, negative three, negative one and 111. We can solve this determinant which would give us negative 30. We can do the same for A B by replacing our Miller row with 74 and 12. You would get 36 91 12 and 111. This is permanent and the calculator is negative 2 40. Finally, we can use DC which will replace our final column with 12. That's 36 N nine N three four and one eight of 1 12. That determinant results in negative 5 70. We can now find out A B and C A will be terminate A over D B will be determinant B over D and C is determinant C over C or D rather RT A was negative 30. RT is negative 30 meaning are A equals one R B is negative 2 40 divided by D negative B will then equals eight. Finally, for C data, the 5 70 divided by negative gives us 19. Our final equation then is why equals X squared plus eight X plus 19. This means the answer to our problem is answer D OK. I hope to help you solve the problem. Thank you for watching. Goodbye.

Struggling with College Algebra?

Use a system of equations to solve each problem. See Example 8. Find an equation of the parabola y = ax^2 + bx + c that passes through the points (2, 3), (-1, 0), and (-2, 2).

Watch next

Two Variable Systems of Linear Equations practice set