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# help!!!!

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

Jul 18, 2020

#1
+28021
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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 !

Jul 18, 2020
#2
+28021
+1

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

Jul 18, 2020
#3
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Thank you!!

Guest Jul 18, 2020
#4
+28021
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I hope you actually looked at the method of solution and understand !

You are welcome .....

ElectricPavlov  Jul 18, 2020