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

Answer:

$a+b+c=10$

Step-by-step 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:

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