# Quadratic equation and two points.

I need to solve a quadratic equation (actually I need to explain it to my kid), but I get stuck in the middle and would be grateful, for any pointers into the right direction.

$$y=ax^2+bx-1$$

with two points on its parabola: $$P_1=(-2,1)$$ and $$P_2=(3,1)$$. I need to find $$a$$ and $$b$$.

Inserting the coordinates, I get \begin{align*} 1 &= 4a-2b-1\\ 1 &= 9a+3b-1 \end{align*} which yields $$2 = 13a+b-2.$$ This gives

$$b = -13a .$$

Now $$-13$$ is a strange number for a schoolbook example. And using Geogebra I found out that $$a$$ and $$b$$ must be $$1/3$$. In all the other questions around it, I see full number quotients and usually either $$a$$ or $$b$$ cancel themselves out, so it is easy to find out the result by substituting.

I am unsure about how to continue because I have the feeling I am missing something here.

• You cancelled out the $2$ and $-2$, but they have opposite signs. Also, what you get is simply a system of linear equations. The easiest way to solve this would be elimination. Multiply the top equation by $3$, the bottom by $2$, and since the $b$'s are now opposite in coefficient, you can add the two equations and obtain $a$. Plug in $a$ in either equation to get $b$, and you're done. – KM101 May 20 at 16:53
• Try taking the difference between your first two equations instead. It works out much nicer. – Michael Seifert May 20 at 16:55
• I've edited your post; please make sure it still has what you wanted it to have. – Clayton May 20 at 16:58

You will get the system $$2=4a-2b$$ and $$2=9a+3b$$ simplifying gives $$1=2a-b$$ $$2=9a+3b$$ Multiplying the first equation by $$3$$ and adding to the second we get $$5=15a$$ so $$a=\frac{1}{3}$$

Your mistake is $$b=-13a$$ which should have been $$b=-13a+4$$

There are various ways to deal with your simultaneous equations - you do need to be careful about signs. You can rearrange them as:

$$4a-2b=2$$

$$9a+3b=2$$

Now to eliminate $$b$$ multiply the first equation by $$3$$ (the coefficient of $$b$$ from the second) and the second equation by $$2$$ (minus the coefficient of $$b$$ from the first equation). You get:

$$12a-6b=6$$

$$18a+6b=4$$

And if you add these you will get $$30a=10$$ or $$a=\frac 13$$. Substituting this in the second equation gives $$3+3b=2$$ or $$b=-\frac 13$$

This is a simple mechanical method - there are often short cuts. Intelligent choices simplify arithmetic. I chose to multiply by the coefficients of $$b$$ rather than eliminating $$a$$ because these were smaller. I chose to substitute back into the second equation because $$9a$$ came out an integer.

As $$P_1$$ and $$P_2$$ share the same second coordinate, the first coordinate of the parabola's vertex is the mean of their first coordinates, namely $$(-2+3)/2=1/2$$, on the other hand it is $$-\frac{b}{2a}$$. From here $$b=-a$$. Now plug in the coordinates of one point in $$y=ax^2-ax-1$$.