System of equations - what are the values of the coefficients of a quadratic parabola?

In the process of relearning the mathematical basics I'm stumbling over this problem:

A quadratic parabola $y = ax^2 + bx + c$ goes through the points A(1/2), B(3/7) and C(-1/1). What are the values of the coefficients $a$, $b$ and $c$?

This is a problem presented in the section about "Systems of Equalities", but I don't have the slightest idea, how to use the coordinates to calculate the values of the coefficients.

How can I solve this problem with a system of equalities?

• You've written "A(1/2)". That is not a point. A point has an $x$-coordinate and a $y$-coordinate. Did you perhaps mean to write "A(1,2)"?
– MJD
May 15 '12 at 14:34
• Maybe it's just European convention or something else. My book does use A(1/2) to tell me the x- and y-coordinates. May 15 '12 at 14:48
• Thanks for the correction. I had not seen that notation before.
– MJD
May 15 '12 at 15:05

Assuming Mark Dominus' interpretation of what you wrote, the fact that the parabola passes through the point $(1,2)$ tells you that if you plug in $x = 1$ and $y = 2$ into the general form of the parabola, the equation on the coefficients must be satisfied for your particular parabola. In this case, you get $2 = a(1)^2 + b(1) + c$ or $a + b + c = 2$. Similar reasoning with the other two points will produce for you two more equations involving $a, b, c$. Thus, you will obtain a system of three equations in three unknowns. Have you learned the techniques for solving such systems?

• Yeah, I know how to solve a system of equations. Thanks for your answer, it saved me some valuable time! May 15 '12 at 14:50

First use the fact that the curve passes through the point $(1,2)$. That says that when $x=1$, we have $y=2$. So substitute $x=1$ in the equation $y=ax^2+bx+c$. We get $$2=a+b+c.$$

Similarly, because $(3,7)$ is on the curve, we have $$7=9a+3b+c.$$ And finally, the third point tells us that $$1=a-b+c.$$ We now have $3$ linear equations in the $3$ unknowns $a$, $b$ and $c$.

From the two equations $9a+3b+c=7$ and $a+b+c=2$, we obtain, by subtraction, that $8a+2b=5$.

From the two equations $9a+3b+c=7$ and $a-b+c=1$, we obtain, by subtraction, that $8a+4b=6$. (I deliberately did not subtract the third from the first, that would have made things too easy!)

We have "eliminated" $c$, and we have two equations in the variable $a$ and $b$. Now we will "eliminate" $b$, which again is not the clever thing to do.

So recall we have $8a+2b=5$ and $8a+4b=6$. Multiply both sides of the first equation by $2$. We get $16a+4b=10$. By subtraction, using $8a+4b=6$, we get $8a=4$, and therefore $a=1/2$. Then from $8a+2b=5$ we get $4+2b=5$ and therefore $b=1/2$. Finally, from $a+b+c=2$ we get that $c=1$.

• Thanks for laying it out. At least I got to compare my results with yours. :) May 15 '12 at 15:04
• @MiroslavCetojevic: The quick way is to start by subtracting third from first. We immediately get $2b=1$. May 15 '12 at 15:42