Question

The graph of the quadratic function y = ax^2 + bx + c is given. Find a + b + c.

Answer 1

Answer:

$\frac{145}{24}\approx 6$***

Answers may vary depending on the person reading the graph.

It is hard to tell what they think crosses nicely.

For example, I decided that the graph crossed nicely at the following points:

$(0,3)$

$(2,8)$

$(6,5)$

Step-by-step explanation:

I see that the graph crosses at $(0,3)$,$(2,8)$, and $(6,5)$.

The equation $y=ax^2+bx+c$ tells us the $y$-intercept is $c$ since when $x=0$ we have $y=a(0)^2+b(0)+c=0+0+c=c$.

The $y$-intercept for our graph is $3$. Therefore, $c=3$.

So far we have the equation:

$y=ax^2+bx+3$.

Let's enter the other points creating a system of linear equations to solve:

$8=a(2)^2+b(2)+3$

$5=a(6)^2+b(6)+3$

Let's simplify:

$8=4a+2b+3$

$5=36a+6b+3$

Let's subtract 3 on both sides:

$5=4a+2b$

$2=36a+6b$

I choose to solve the system by elimination.

Let's multiply the top equation by -3:

$-15=-12a-6b$

$2=36a+6b$

Now adding the equations results in:

$-13=24a$

Divide both sides by 24:

$\frac{-13}{24}=a$

Now using one of the equations we can find $b$:

$5=4a+2b$ with $a=\frac{-13}{24}$

$5=4\frac{-13}{24}+2b$

$5=\frac{-13}{6}+2b$

Add 13/6 on both sides:

$5+\frac{13}{6}=2b$

$\frac{30+13}{6}=2b$

$\frac{43}{6}=2b$

Divide both sides by 2:

$\frac{43}{2(6)}=b$

$\frac{43}{12}=b$

So we have the equation:

$y=\frac{-13}{24}x^2+\frac{43}{12}+3$

Let's evaluate $a+b+c$ now:

$\frac{-13}{24}+\frac{43}{12}+3$

$\frac{-13+86+72}{24}$

$\frac{73+72}{24}$

$\frac{145}{24}$