Answer:
[tex]a+b+c=10[/tex]
Step-by-step explanation:
We are given that the graph of the equation:
[tex]y=ax^2+bx+c[/tex]
Passes through the three points (0, 5), (1, 10), and (2, 19).
And we want to find the value of (a + b + c).
First, since the graph passes through (0, 5), its y-intercept or c is 5. Hence:
[tex]y=ax^2+bx+5[/tex]
Next, since the graph passes through (1, 10), when x = 1, y = 10. Substitute:
[tex](10)=a(1)^2+b(1)+5[/tex]
Simplify:
[tex]5=a+b[/tex]
The point (2, 19) tells us that when x = 2, y = 19. Substitute:
[tex](19)=a(2)^2+b(2)+5[/tex]
Simplify:
[tex]14=4a+2b[/tex]
This yields a system of equations:
[tex]\begin{cases} 5 = a + b \\ 14 = 4a + 2b\end{cases}[/tex]
Solve the system. We can do so using elimination (or any other method you prefer). Multiply the first equation by negative two:
[tex]-10=-2a-2b[/tex]
Add the two equations together:
[tex](-10)+(14)=(-2a+4a)+(-2b+2b)[/tex]
Combine like terms:
[tex]4 = 2a[/tex]
Hence:
[tex]a=2[/tex]
Using the first equation:
[tex]5=(2)+b\Rightarrow b=3[/tex]
Therefore, our equation is:
[tex]y=2x^2+3x+5[/tex]
Thus, the value of (a + b + c) will be:
[tex]a+b+c = (2) + (3) + (5) = 10[/tex]
