how do i determine the equation of the quadratic function that passes through (-4,5) if its zeroes are 2+√3 and 2-√3

Question

campbell_st · Answer

start with the perfect square  (x -2)^2 = 3   or (x - 2)^2 - 3 = 0
 
because when the solve for x 

$$(x -2) = \pm \sqrt{3}$$

$$x = 2\pm \sqrt{3}$$

then you have the equation 

y = a((x - 2)^2 - 3)

y = a(x^2 - 4x + 1)

substitute the point to find a

5 = a((-4)^2 - 4(-4) + 1)

5 = a(16 + 16 + 1)

a = 5/33

equation y = 5/33(x^2 - 4x + 1)

anonymous · Answer

would i do the same if i'm given the x-intercepts √7 and -√7 and the parabola passes through the point (-5,3)

campbell_st · Answer

something similar that seems easiest to me..... its a solution I worked through, someone may have a more elegant method

anonymous · Answer

thanks!