The graph of y=ax^2+bx+c passes through points (0,5), (1,10), and (2,19). Find a+b+c.
+36916 Use the first point , 0,5 to find the value of 'c'
then use the second set of points to find an equation with 'a' and 'b' in it
use the third set of points to find ANOTHER equation with 'a' and 'b' in it
then use the two equations to solve for 'a' and 'b' via elimination or substitution.
Re-post if you need more help.... Go !
+36916 at point 0,5
5 = a(02) + b (0) + c
5 = c
Now sub in the point 1,10
10 = a(1^2) + b(1) + 5
10 = a + b + 5
5 = a + b
Now put in the point 2,19
19 = a (2^2) + b(2) + 5
19 = 4a + 2b + 5
14 = 4a+ 2b
Now you have two (RED) equations and two unknowns (a and b) solve for a and b
a+b = 5
4a+ 2b = 14 I'll solve by 'elimination'
.... multiply the first equation by -2 then add to the second equation to get :
2a = 4
a = 2 and since a+b = 5 b = 3 and c = 5
+36916 I hope you actually looked at the method of solution and understand !
You are welcome .....
