How do i solve this quadratic function.

Question

Find a quadratic function in standard form for each set of points.

(0,3), (1,-4), (2,-9), (2,2)

 

I know how to do a 3 point equation but I'm stumped here, can someone explain to me how to solve this step by step. Thanks! 

Answer 1

Just choose any three of the points and get the associated parabola's equation. Then check to see that the unchosen point also satisfies the equation.

Answer 2

Notice that (0, 3) is the y-intercept.  In a quadratic equation     y=ax²+bx+c  , c is the y-intercept.  Therefore, we can say the general equation of this parabola is

 

y = ax² + bx + 3

 

 

We use this equation to form 3 equations that will be used to solve.  The other 3-points will correspond to those equations.  For each point, plug in the x and y value into the general parabola.

 

For the point  (1, -4):

 

a(1)² + b(1) + 3 = -4

a + b = -7                          eq1

 

 

For the point (2, -9):

 

a(2)² + b(2) + 3 = -9

4a + 2b = -12                      eq2

 

 

For the point (2, 2):

 

a(2)² + b(2) + 3 = 2

4a + 2b = -1                         eq3

 

 

The three bolded equations are your system of equations to solve for a and b.  Use substitution/elimination method to solve.

 

 

But, here is the problem.  The last two points (2, -9) and (2, 2) do not form a function.  This is because these two points have y values that share a common  x value.  Because of this, you get a vertical line when you connect them, and not a parabola.

 

You need to revise those points, and follow the technique I have shown here.