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[SOLVED] find quadratic equation given 3 points

karush

Well-known member
Jan 31, 2012
2,718
how do you find the quadratic equation given (-3,0) (3,0) and (0,-4)

thus y=k(x-3)(x+3) for the zero's but how do you find k or any other better method of finding the equation

thanks much(Cool)
 
Last edited:

Jameson

Administrator
Staff member
Jan 26, 2012
4,041
This can be done by looking at the general form of a quadratic equation: \(\displaystyle y=ax^2+bx+c\). We need to solve for a,b and c in order to write our equation and we have three points so we can do this through substitution. First use (0,-4) for (x,y) and you get \(\displaystyle -4=a(0)^2+b(0)+c\) which means that c=-4. Now use the other two points the same way and you will have to solve a two variable system of equations for a and b. Once you have a,b and c you have your answer.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,703
how do you find the quadratic equation given (-3,0) (3,0) and (0,-4)

thus y=k(x-3)(x+3) for the zero's but how do you find k or any other better method of finding the equation
Starting from y=k(x-3)(x+3) (which gives the value 0 when x = 3 or –3), all you need to do is to put x=0 to see that y = –9k when x=0. But you want y to be –4 when x=0. Therefore –9k = –4. So k = 4/9, and $y = \frac49(x-3)(x+3).$
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
how do you find the quadratic equation given (-3,0) (3,0) and (0,-4)
Another way (if you have covered the Lagrange Interpolation Polynomial):

$y=0L_1+0L_2-4L_3=-4\dfrac{(x+3)(x-3)}{(0+3)(0-3)}=\dfrac{4}{9}(x+3)(x-3)$
 

karush

Well-known member
Jan 31, 2012
2,718
Another way (if you have covered the Lagrange Interpolation Polynomial):

$y=0L_1+0L_2-4L_3=-4\dfrac{(x+3)(x-3)}{(0+3)(0-3)}=\dfrac{4}{9}(x+3)(x-3)$
no have not heard of it. looks valuable tho so will look it up thanks