How to find a quadratic's coefficients from two points and a constant term

Key takeaway: Once the constant term is fixed, the graph passing through two points gives you two linear equations in $a$ and $b$. The algebra is routine, but the setup has to be exact.

We are given

$y=ax^2+bx+13$

and the graph passes through $(-3,-23)$ and $(4,5)$.

Step 1: Substitute $(-3,-23)$

$-23=a(-3)^2+b(-3)+13$ $-23=9a-3b+13$ $-36=9a-3b$

So we get

$9a-3b=-36$

Step 2: Substitute $(4,5)$

$5=a(4)^2+b(4)+13$ $5=16a+4b+13$ $-8=16a+4b$

So we get

$16a+4b=-8$

Step 3: Solve the system

Divide the first equation by 3:

$3a-b=-12$

Divide the second equation by 4:

$4a+b=-2$

Now add the equations:

$7a=-14$ $a=-2$

Substitute into $4a+b=-2$:

$4(-2)+b=-2$ $-8+b=-2$ $b=6$

Final answer

$\boxed{a=-2,\ b=6}$

A quick check is worth doing:

$y=-2x^2+6x+13$

At $x=-3$:

$y=-2(9)-18+13=-23$

At $x=4$:

$y=-2(16)+24+13=5$

Both points fit.

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Pitfalls the pros know 👇 Do not mix up the two substitutions. A small sign error in $-3b$ or $-23-13$ can throw the whole system off. Also, remember the constant term is already built in as $13$, so there is no need to solve for it.

What if the problem changes? If the question had used three points instead of two, you could still set up a system, but the extra point would act as a check. If the constant term were unknown as well, you would need three independent equations to solve for $a$, $b$, and $c$.

Tags: system of equations, quadratic coefficients, substitution