
1) Given the intercepts and (5,0) and point (7,-16) let's plug it in
Let's set a system of linear equations to find a:
49a +7b +c
9a -3b +c
25a +5b +c
49a +7b +c=16
9a -3b +c=0 x (-1) To eliminate c
49a +7b +c =16
-9a +3b -c=0
49a +7b = 16 Adding both equations simultaneously
-9a +3b = 0
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40a +10b= 16
2) Now let's operate the other two equations, eliminating the c variable:
25a +5b +c =0
9a -3b +c= 16 x(-1)
25a +5b+c =0
-9a +3b -c=0
--------------------
16a +8b = 0
Now let's solve the system for a and b. Let's eliminate b:
40a +10b= 16 x 4
16a +8b = 0 x-5
160a +40b = 64
-80a -40b = 0
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80a = 64
a = 4/5
3) Now we can write this quadratic equation since the x-intercepts are x=-3 and x= 5
