• Apr 17th 2013, 05:47 PM
Fratricide
1) Find the equation of the quadratic which passes through the points with coordinates: (-2, -1), (1, 2), (3, -16).

2) An arch on the top of a door is parabolic in shape. The point A is 3.1 m above the bottom of the door. The equation y = ax2 + bx + c can be used to describe the arch. Find the values of a, b and c.

Attachment 28000
(Sorry about the small picture, if it needs to be larger I can re-upload it.)

• Apr 17th 2013, 05:55 PM
Prove It
1. Substitute the three points into the equation $\displaystyle \displaystyle y = a\,x^2 + b\,x + c$ to give three equations in three unknowns. Solve them simultaneously for a,b,c.

2. Surely you can tell me what the coordinate of the turning point is. You can also see (0,0) and (1.5, 0) lie on the curve. So either do the same thing as in part 1, or substitute the turning point into the turning point form of the quadratic $\displaystyle \displaystyle y = a\left( x - h \right) ^2 + k$ and then use one of the other points to help find a.
• Apr 17th 2013, 05:59 PM
MarkFL
For the first problem, let the quadratic you are seeking be:

$\displaystyle f(x)=ax^2+bx+c$

and use the 3 given points to create a 3X3 linear system of equations.

Can you state this system?
• Apr 17th 2013, 08:51 PM
Fratricide
So for the first problem the three equations would be:

-1 = 4a - 2b + c
2 = a + b + c
-16 = 9a + 3b + c

How exactly would I solve them simultaneously? I've never simultaneously solved equations with more than two variables before.
• Apr 17th 2013, 09:00 PM
MarkFL
I would suggest multiplying the second equation by -4, and add it to the first to eliminate a. Then you have a 2X2 system, which you know how to solve. Then once you have b and c, substitute them into any of your 3 equations to find a.
• Apr 17th 2013, 09:00 PM
Prove It
First of all, I'd put the second equation first (it's just easier with its coefficients).

\displaystyle \displaystyle \begin{align*} 2 &= a + b + c \\ -1 &= 4a - 2b + c \\ -16 &= 9a + 3b + c \end{align*}

The process is the same as if it was just two equations in two unknowns. Try to eliminate c from the two bottom equations. Then after you've done that, try to eliminate b from the last equation. This will give you an equation you can solve for a, back substitute and solve for b, back substitute a and b to solve for c.
• Apr 19th 2013, 05:07 PM
Fratricide
Okay. So I completed the second question without any troubles. As for the second equation, I multiplied 2 = a + b + c by -4 and added it to -1 = 4a - 2b + c to eliminate a. The resulting equation was -9 = -6b - 3c. I'm having trouble identifying the 2x2 system that I should now have in order to continue the question.

I'm terribly sick today and apologise for any ignorant questions I may ask.
• Apr 19th 2013, 06:06 PM
MarkFL
Quote:

Originally Posted by Fratricide
Okay. So I completed the second question without any troubles. As for the second equation, I multiplied 2 = a + b + c by -4 and added it to -1 = 4a - 2b + c to eliminate a. The resulting equation was -9 = -6b - 3c. I'm having trouble identifying the 2x2 system that I should now have in order to continue the question.

I'm terribly sick today and apologise for any ignorant questions I may ask.

My reply above was incomplete, so this is my fault. Let's look at your 3X3 system:

(1) $\displaystyle -1=4a-2b+c$

(2) $\displaystyle 2=a+b+c$

(3) $\displaystyle -16=9a+3b+c$

Multiplying (2) by -4 and adding to (1) we obtain (which you correctly found):

$\displaystyle -9=-6b-3c$ or by dividing through by -3:

(4) $\displaystyle 2b+c=3$

Now this is one of the equations for the 2X2 system. Can you see how you can eliminate $\displaystyle a$ from equations (2) and (3)? What should you multiply equation (2) by so that when you add it to (3), $\displaystyle a$ is eliminated?
• Apr 20th 2013, 04:56 PM
Fratricide