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 $y = \frac49(x-3)(x+3).$karush said: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
karush said:how do you find the quadratic equation given (-3,0) (3,0) and (0,-4)
Fernando Revilla said:Another way (if you have covered the Lagrange Interpolation Polynomial):
$y=0L_1+0L_2-4L_3=-4\dfrac{(x+3)(x-3)}{(0+3)(0-3)}=\dfrac{4}{9}(x+3)(x-3)$
A quadratic equation is an algebraic expression that contains a variable raised to the second power. It is typically written in the form of ax^2 + bx + c = 0, where a, b, and c are constants and x is the variable.
To find the quadratic equation given 3 points, you need to set up a system of equations using the coordinates of the points. This will result in a system of 3 equations with 3 variables, which can then be solved using algebraic methods to find the values of a, b, and c in the quadratic equation.
You will need the coordinates of the 3 points on the quadratic curve. This includes the x and y values for each point. It is also helpful to have a basic understanding of how to solve systems of equations using algebraic methods.
Yes, there are typically multiple quadratic equations that can pass through 3 given points. This is because there are multiple ways to solve the system of equations and find values for a, b, and c. However, there is usually one quadratic equation that is the simplest and most commonly used.
Yes, it is possible to find a quadratic equation given more than 3 points. However, the more points you have, the more complex the system of equations will be and the more difficult it will be to solve. It is recommended to use a computer or calculator to solve for the quadratic equation when given more than 3 points.