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so I'm doing this math problem for my Calculus I course in college. Here is a screenshot of the problem: Graph 1 (click here to view); the prompt is "Find an expression for the quadratic function whose graph is shown."

It is obvious from the graph that the three co-ordinate points are $(1, -6.5)$, $(0,1)$ and $(-4,11)$.

Now, one way of doing is to plug in the values into the equation "$y=ax^2+bx+c$". However, I tried a different method, and I did not get the correct answer. I want to know if my method is correct and if not, what caveat(s) are there.

The formula I am using is "$y=-p(x-q)^2+r$", which is basically derived from the basic quadratic formula "$y=x^2$", with $p$ being the shrink factor, $q$ being the horizontal translation factor, $r$ being the vertical translation factor, and the negative sign ("$-$") before $p$ coming because, as is evident from the picture above, $g(x)$ is a negative function.

Plugging in my three data points into the above function, I get three equations:

$(\text{Eq. }1) \Rightarrow \quad 11=-p(-4-q)^2+r \quad \Rightarrow \quad 11=-16p-8qp-q^2p+r \\(\text{Eq. }2) \Rightarrow \quad -6.5=p(1-q)^2+r \quad \Rightarrow \quad -6.5=-p+2qp-q^2p+r\\(\text{Eq. }3) \Rightarrow \quad 1=-p(q)^2 +r \qquad \quad \Rightarrow \quad 1=-q^2p+r$

Next, separately, I am going to subtract each equation from the other, i.e, first I will do $(1)-(2)$, then $(1)-(3)$, then finally $(2)-(3)$, thus forming three more equations (Eqs. 4, 5 and 6):

$(\text{Eq. }4) \Rightarrow \quad (1)-(2)\text{ gives } \quad 17.5=-17p-10qp\\(\text{Eq. }5) \Rightarrow \quad (1)-(3)\text{ gives } \quad 10=-16p-8qp\\(\text{Eq. }6) \Rightarrow \quad (2)-(3)\text{ gives } \quad -7.5=-p+2qp$

Solving each of the three above equations for $p$, I get three different equations (Eqs. 7, 8 and 9) for $p$ in terms of $q$:

$(\text{Eq. }7) \Rightarrow \quad (4)\text{ gives } \quad p=\frac{-17.5}{10q+17}\\(\text{Eq. }8) \Rightarrow \quad (5)\text{ gives } \quad p=\frac{-5}{4q+8}\\(\text{Eq. }9) \Rightarrow \quad (6)\text{ gives } \quad p=\frac{-7.5}{2q-1} $

Here is where the problem arises. If the steps above are all correct, then by equating the above three equations 7, 8 and 9 with each other, I should get a consistent value for $q$. I do not.

$\text{Equating }(7)\text{ and }(8)\text{ gives me } q=-2.75\\\text{Equating }(8)\text{ and }(9)\text{ gives me } q=-3.25\\\text{Equating }(7)\text{ and }(9)\text{ gives me } q=-3.625 $

This does not make any sense to me. Again, I could solve the problem using the simple quadratic equation I mentioned above ($y=ax^2+bx+c$), but I will always wonder where I went wrong when solving it using this method.

Obviously, this method looks right to me. Please review it, as well as the individual calculations if you must (though I am pretty sure there are no silly arithmetic errors or such, as I have proofread it over and over), and tell me what mistake(s) I am making. Of course, review the method itself (including the simultaneous equations as well as the plugging in of values).

To whoever takes the time to help solve this problem of mine, a big THANK YOU in advance! As you can see by the detail I put into explaining this problem here and ensuring that all the notation is correct, it would really mean a lot to me if you could get to the bottom of this problem. Thanks!

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Eq. 4 is wrong. It should be $-15p$, not $-17p$. There is a typo in Eq. 2, but that does not affect the calculation.

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  • $\begingroup$ I see what you're saying André, thanks for that. I think you mean Eq. 4 is wrong, not Eq. 5. But yeah, it should be -15p and not -17p. Also, I see the typo in Eq. 2. Thanks $\endgroup$ – Karan Erry Jan 28 '16 at 4:36
  • $\begingroup$ You are welcome. Yes, it is 4. $\endgroup$ – André Nicolas Jan 28 '16 at 4:38
  • $\begingroup$ You did a very energetic series of calculations, which most students would have feared to undertake. Congratulations. Minus sign errors are frequent, I have made many more of them than you (I am substantially older.) An easier way is a variant of the $a,b,c$ you mentioned. Since the parabola passes through $(0,1)$, the constant term $c$ is $1$, and we are looking at $y=ax^2+bx+1$, only two variables. $\endgroup$ – André Nicolas Jan 28 '16 at 16:31
  • $\begingroup$ That's really encouraging André, thanks for that. Yes I know what you mean -- these arithmetic errors are much less than I used to make a year ago, so that's encouraging. Leaving out the arithmetic and actual values for a moment, is my method correct? In other words, if I did all the calculations correctly, should I get consistent values for q, or is there something in my methodology that looks incorrect? $\endgroup$ – Karan Erry Jan 28 '16 at 17:52
  • $\begingroup$ Your method is correct. $\endgroup$ – André Nicolas Jan 28 '16 at 17:55

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