Use a system of equations to find the equation of the parabola y = ax^2 + bx + c that passes through the points. Solve the system using matrices. Use a graphing utility to verify your result.
The points are (1,6), (4,3), (5,-2)
@reiny thanks but how did you get 16a in 3 = 16a + 4b + c and 25a in -2 = 25a + 5b + c
using (1,6) ---> 6 = a + b + c
using (4,3) ---> 3 = 16a + 4b + c
using (5,-2) ---> -2 = 25a + 5b + c
The 2nd minus the first: 15a + 3b = -3 or 5a + b = -1
the 3rd minus the 2nd: 9a + b = -5
Now subtract those last two:
4a = -4
a = -1
sub that into 5a + b = -1 to get b
then go into the very first one to get c
your equation is y = ax^2 + bx + c
so in (4,3), x = 4, and y = 3
so 3 = a(4^2) + b(4) + c
same for the other points
thanks, now i get it
Why did the parabola go to the therapist?
Because it had too many ups and downs!
Let's find the equation of the parabola that passes through the points (1,6), (4,3), and (5,-2) using a system of equations and matrices.
We can start by substituting the x and y values of the given points into the equation of the parabola: y = ax^2 + bx + c
For the point (1,6):
6 = a(1)^2 + b(1) + c => 6 = a + b + c
For the point (4,3):
3 = a(4)^2 + b(4) + c => 3 = 16a + 4b + c
For the point (5,-2):
-2 = a(5)^2 + b(5) + c => -2 = 25a + 5b + c
Now we have a system of three equations:
1) 6 = a + b + c
2) 3 = 16a + 4b + c
3) -2 = 25a + 5b + c
To solve this system using matrices, we can write the augmented matrix:
[ 1 1 1 | 6 ]
[ 16 4 1 | 3 ]
[ 25 5 1 | -2 ]
Using matrix operations, we can perform row operations to reduce the matrix into row-echelon form:
[ 1 1 1 | 6 ]
[ 0 -12 -15 | -93 ]
[ 0 0 -24 | -111 ]
From this row-echelon matrix, we can solve for the values of a, b, and c as follows:
c = -111 / -24 = 111/24 = 37/8
b = -93 / -12 = 31/4
a = 6 - (31/4) - (37/8) = -1/8
Therefore, the equation of the parabola that passes through the given points is:
y = (-1/8)x^2 + (31/4)x + 37/8
You can use a graphing utility to verify the graph of this equation. Enjoy exploring the funny world of parabolas!
To find the equation of the parabola that passes through the given points, we can form a system of equations using the coordinates of these points.
The general form of a quadratic equation is: y = ax^2 + bx + c
To create the system of equations, substitute the x and y values of each point into the equation, and solve for the coefficients a, b, and c.
Using the given points:
(1,6): 6 = a(1)^2 + b(1) + c
(4,3): 3 = a(4)^2 + b(4) + c
(5,-2): -2 = a(5)^2 + b(5) + c
Rewriting these equations in standard form:
1a + 1b + c = 6 (Equation 1)
16a + 4b + c = 3 (Equation 2)
25a + 5b + c = -2 (Equation 3)
Now, let's form a matrix equation to solve the system using matrices:
| 1 1 1 | | a | | 6 |
| 16 4 1 | * | b | = | 3 |
| 25 5 1 | | c | | -2 |
To solve this matrix equation, we can use matrix operations to find the values of a, b, and c.
Using a graphing utility, we can input the matrix equation, and use the matrix solver function to find the values of a, b, and c.
After solving the matrix equation, we get:
a = -1/10
b = 3/5
c = 43/10
Therefore, the equation of the parabola that passes through the given points is:
y = (-1/10)x^2 + (3/5)x + 43/10
To verify this result, graph the equation using a graphing utility and input the given points. You should see that the parabola passes through the points (1,6), (4,3), and (5,-2).