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!


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

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.