MHB Finding equation of parabola with 3 points given

[phoenixcat]

Hello, I am supposed to find the equation of a parabola with the points (8,10)(11,10)(10,20/3). I have tried putting these values into y=ax^2+bx+c, but get different answers each time, like c=-160, which is not right! A step by step explanation would be greatly appreciated, as I am very unsure what I'm doing wrong

 

[MarkFL]

Your technique is good and will work:

$$a(8)^2+b(8)+c=10$$

$$a(11)^2+b(11)+c=10$$

$$a(10)^2+b(10)+c=\frac{20}{3}$$

We can rewrite this system as:

$$64a+8b+c=10$$

$$121a+11b+c=10$$

$$300a+30b+3c=20$$

Now, here is what I suggest...Subtract the first equation from the second and subtract the third equation from 3 times the second to eliminate $c$ and obtain a 2X2 system in $a$ and $b$. The using this system, subtract the first from the second to eliminate $b$ and solve for $a$. Use this value of $a$ in either of these two equations to find $b$ and then use the values of $a$ and $b$ in the third equation above to find $c$.

What do you find?

 

[phoenixcat]

I end up with y=5/3x^2 -95/3x +470/3.. I think it should work, but I was also given a not-to-scale diagram with the question, and it shows the TP to have a smaller y value, the TP is above the x-axis and below 10, it's confusing, hmm

 

[MarkFL]

Those are the correct values, and I have verified the parabola passes through the given points. Here is a plot:

View attachment 2244

 

[phoenixcat]

Haha, accidentally thought for a second(more than a few seconds) finding the Y intercept was finding the y value of the tp (Giggle) thanks for the help, will try applying this to future situations!