How do you write a quadratic equation with x-intercepts: -4,3 ; point: (-5,16)?

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Answer 1 · 2016-09-11T13:06:14

#y=2x^2+2x-24#

Use the form #y=k(x-a)(x-b)#
where k is a constant, and a and b are the zeros (x coordinate of the x-intercepts).

#a=-4# and #b=3#

#y=k(x+4)(x-3)#

#y=k(x^2 +4x-3x-12)#

#y=k(x^2+x-12)#

To find k, plug in the given point #(-5,16)#.

#16=k((-5)^2+(-5)-12)#
#16=k(25-5-12)#
#16=k(8)#
#k=2#

Plug k into the equation found above.
#y=2(x^2+x-12)#

#y=2x^2+2x+24#