The other tutor gave you a way that will work ONLY if you are given
the intercepts. But you should do it a more general way, because
the three points you are given will not always be intercepts. For
instance you may be given the points (7,6), (-3,24) and (12,42).
Here's what you should do instead. You must start with
y = ax˛ + bx + c
Then you substitute in each of the three points, the x-coordinate
for x, and the y-coordinate for y.
Substituting (2,0)
y = ax˛ + bx + c
0 = a(2)˛ + b(2) + c
0 = 4a + 2b + c
Substituting (5,0)
y = ax˛ + bx + c
0 = a(5)˛ + b(5) + c
0 = 25a + 5b + c
Substituting (0,6)
y = ax˛ + bx + c
6 = a(0)˛ + b(0) + c
6 = 0 + 0 + c
6 = c
Now you have three equations:
0 = 4a + 2b + c
0 = 25a + 5b + c
6 = c
Substitute 6 for c in the first two:
0 = 4a + 2b + 6
0 = 25a + 5b + 6
Eliminate the b's by multiplying the first equation by -5
and the second equation by 2
0 = -20a - 10b - 30
0 = 50a + 10b + 12
-------------------
0 = 30a - 18
18 = 30a
= a
= a
Substitute for a in
0 = 25a + 5b + 6
0 = 25 + 5b + 6
0 = 15 + 5b + 6
0 = 21 + 5b
-21 = 5b
= b
Now substitute for a, = b, and 6 for c in
y = ax˛ + bx + c
and get:
This method will always work whether the points given happen to be
intercepts or not.
Edwin
Question 909255
The other tutor gave you a way that will work ONLY if you are given
the intercepts. But you should do it a more general way, because
the three points you are given will not always be intercepts. For
instance you may be given the points (7,6), (-3,24) and (12,42).
Here's what you should do instead. You must start with
y = ax˛ + bx + c
Then you substitute in each of the three points, the x-coordinate
for x, and the y-coordinate for y.
Substituting (2,0)
y = ax˛ + bx + c
0 = a(2)˛ + b(2) + c
0 = 4a + 2b + c
Substituting (5,0)
y = ax˛ + bx + c
0 = a(5)˛ + b(5) + c
0 = 25a + 5b + c
Substituting (0,6)
y = ax˛ + bx + c
6 = a(0)˛ + b(0) + c
6 = 0 + 0 + c
6 = c
Now you have three equations:
0 = 4a + 2b + c
0 = 25a + 5b + c
6 = c
Substitute 6 for c in the first two:
0 = 4a + 2b + 6
0 = 25a + 5b + 6
Eliminate the b's by multiplying the first equation by -5
and the second equation by 2
0 = -20a - 10b - 30
0 = 50a + 10b + 12
-------------------
0 = 30a - 18
18 = 30a
= a
= a
Substitute for a in
0 = 25a + 5b + 6
0 = 25 + 5b + 6
0 = 15 + 5b + 6
0 = 21 + 5b
-21 = 5b
= b
Now substitute for a, = b, and 6 for c in
y = ax˛ + bx + c
and get:
This method will always work whether the points given happen to be
intercepts or not.
Edwin