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

In the procedure of relearning the mathematical essentials I'm stumbling over this problem:

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

This is a trouble offered in the area concerning "Systems of Equalities", yet I do not have the least suggestion, just how to make use of the works with to compute the values of the coefficients.

Just how can I address this trouble with a system of equals rights?

2022-07-25 20:44:03
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Answers: 2

Assuming Mark Dominus' analysis of what you created, the reality that the parabola travels through the factor $(1,2)$ informs you that if you connect in $x = 1$ and also $y = 2$ right into the basic kind of the parabola, the formula on the coefficients have to be pleased for your certain parabola. In this instance, you get $2 = a(1)^2 + b(1) + c$ or $a + b + c = 2$. Comparable thinking with the various other 2 factors will certainly generate for you 2 even more formulas entailing $a, b, c$. Hence, you will certainly get a system of 3 formulas in 3 unknowns. Have you found out the strategies for addressing such systems?

2022-07-25 22:32:18

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 $c=1$.

2022-07-25 22:31:35