Question

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

Answer 1

`a+b+c=10`

Explanation:

We are given that the graph of the equation:

`y=ax^2+bx+c`

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:

`y=ax^2+bx+5`

Next, since the graph passes through (1, 10), when x = 1, y = 10. Substitute:

`(10)=a(1)^2+b(1)+5`

Simplify:

`5=a+b`

The point (2, 19) tells us that when x = 2, y = 19. Substitute:

`(19)=a(2)^2+b(2)+5`

Simplify:

`14=4a+2b`

This yields a system of equations:

`\begin{cases} 5 = a + b \\ 14 = 4a + 2b\end{cases}`

Solve the system. We can do so using elimination (or any other method you prefer). Multiply the first equation by negative two:

`-10=-2a-2b`

Add the two equations together:

`(-10)+(14)=(-2a+4a)+(-2b+2b)`

Combine like terms:

`4 = 2a`

Hence:

`a=2`

Using the first equation:

`5=(2)+b\Rightarrow b=3`

Therefore, our equation is:

`y=2x^2+3x+5`

Thus, the value of (a + b + c) will be:

`a+b+c = (2) + (3) + (5) = 10`