Question 909255
Question 909255
<pre>
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
{{{18/30}}} = a
{{{3/5}}} = a

Substitute {{{3/5}}} for a in

  0 = 25a + 5b + 6
  0 = 25{{{(3/5)}}} + 5b + 6
  0 = 15 + 5b + 6
  0 = 21 + 5b
-21 = 5b
{{{-21/5}}} = b

Now substitute {{{3/5}}} for a, {{{-21/5}}} = b, and 6 for c in

y = ax˛ + bx + c

and get:

{{{y=expr(3/5)x^2-expr(21/5)x+6}}}

This method will always work whether the points given happen to be
intercepts or not.

Edwin</pre>