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In the picture the parabola passes through points $A=(1,3)$ and $B=(5,-5)$, and the line passes through points $C=(2,-2)$ and $D=(-2,-6)$.

Picture

Find the equation of the parabola.


As we can see, the scale of the graph is $2:1$ for $x$-axis and $1:1$ for $y$-axis.

I have found the equation of the parabola by looking at the graph and seeing that the point $(0,0)$ is in the parabola. Hence, we have $3$ points and we are done.

However, I have never used the points on the line. The exercise seems to want to use them in some way, so is there a method in which without "knowing" that the parabola passes through $(0,0)$ we can use the equation of the line?

Thanks!!

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  • $\begingroup$ Maybe observing that the line and the parabola intersect on the $x$ axis, son you can find the point of intersection and then you have the third point you need. $\endgroup$ – tommycautero Mar 10 '19 at 22:27
  • $\begingroup$ @tommycautero thanks for the help! However, it seems to me that we are not being analytic in that way. I would like to know some analytical form that allows to know the equation without "looking" at the picture. $\endgroup$ – manooooh Mar 10 '19 at 22:30
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    $\begingroup$ I don't think it's possible. The points $(2,-2)$ and $(-2.-6)$ determine a unique line, but there are many parabolas passing through the other two points. Without "looking at the picture", there is no connection whatsoever between the line and the parabola. $\endgroup$ – rogerl Mar 10 '19 at 22:44
  • $\begingroup$ @rogerl yeah, that can be a nice explanation for me! Would you like to post it as an answer? $\endgroup$ – manooooh Mar 10 '19 at 22:46
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To use the line to define the parabola we would have to know some relation between the line and the parabola.

From the picture it does look like the line and the parabola intersect at the point $(4,0)$, but of course we are not told this.

If you use the point $(0,0)$ as a third point like you mentioned and calculate the parabola $y=-x^2 +4x$ then $(4,0)$ does indeed lie on the parabola.

Also, the line does indeed intersect the parabola at exactly $(4,0)$

Are you sure there aren't more instructions in the question ?

And are you sure there isn't a part $a$, $b$, $c$, etc ?

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  • $\begingroup$ No, there is no more information. This question is under the title "Functions". $\endgroup$ – manooooh Mar 11 '19 at 0:42

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