There is a unique parabola that goes through the points P = ( 1, 1), Q = (1, 1), and
R=(5,6) that can be described by y=ax^2 +bx+c. Find coefficients a, b, and c.
P and Q are the same point
need 3 DIFFERENT points
oh sorry P=(-1,1)
plug the points into the general equation and solve the system
P...1 = a - b + c
Q...1 = a + b + c
R...6 = 25a + 5c + c
ok...I got b=0, a=5/24, c=19/24 is that right?
looks good
To find the coefficients a, b, and c of a parabola that passes through the given points P, Q, and R, we can set up a system of equations using the coordinates of these points.
Let's start with the equation of the parabola:
y = ax^2 + bx + c
Now, substitute the coordinates of point P = (1, 1) into the equation:
1 = a(1^2) + b(1) + c
1 = a + b + c -- Equation 1
Next, substitute the coordinates of point Q = (1, 1) into the equation:
1 = a(1^2) + b(1) + c
1 = a + b + c -- Equation 2
Finally, substitute the coordinates of point R = (5, 6) into the equation:
6 = a(5^2) + b(5) + c
6 = 25a + 5b + c -- Equation 3
Now, we have a system of three equations with three unknowns (a, b, and c). We can solve this system to find the values of a, b, and c.
Combining Equations 1 and 2, we have:
1 = a + b + c
Subtracting Equation 2 from Equation 3, we get:
6 - 1 = 25a + 5b + c - (a + b + c)
5 = 24a + 4b -- Equation 4
Now, we need to solve Equation 4 with Equation 1 simultaneously. Rearranging Equation 4, we get:
24a = 5 - 4b
Substituting this into Equation 1, we have:
1 = (5 - 4b) + b + c
1 = 5 - 3b + c
Rearranging this equation:
c = -3b - 4 -- Equation 5
Now, substitute Equation 5 into Equation 1:
1 = a + b - 3b - 4
1 = a - 2b - 4
Rearranging this equation:
a = 2b + 5 -- Equation 6
We now have expressions for a, b, and c in terms of b. Substituting Equation 6 and Equation 5 back into Equation 4, we get:
5 = 24(2b + 5) + 4b
5 = 48b + 120 + 4b
5 = 52b + 120
52b = -115
b = -115/52
Substituting this value of b back into Equation 6, we can find a:
a = 2(-115/52) + 5
Simplifying this equation, we have:
a = (-230/52) + (260/52)
a = 30/52
a = 15/26
Finally, substituting the values of a and b into Equation 5, we can find c:
c = -3(-115/52) - 4
Simplifying this equation, we have:
c = (345/52) - 208/52
c = 137/52
Therefore, the coefficients of the parabola that passes through the given points P, Q, and R are:
a = 15/26
b = -115/52
c = 137/52