The graph of y=ax^2+bx+c passes through points (0,5), (1,10), and (2,19). Find a+b+c.

Guest Jul 18, 2020

#1**+1 **

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 !

ElectricPavlov Jul 18, 2020

#2**+1 **

at point 0,5

5 = a(0^{2}) + 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

ElectricPavlov Jul 18, 2020

#4**0 **

I hope you actually looked at the method of solution **and**** understand** !

You are welcome .....

ElectricPavlov
Jul 18, 2020