How do you write a polynomial equation for a graph that passes through point (2, -14) and has x-intercepts at (rad5,0), (-rad5,0), and bounces off x-axis at (6,0)?

1 Answer
Teddy
Jun 27, 2018

#f(x)=7/8(x^2-5)(x-6)^2#

Explanation:

Since we are given that the polynomial has x-intercepts or zeroes at #(+-sqrt5,0)#, we know that the polynomial must have factors #(x-sqrt5)# and #(x+sqrt5)#. In addition, we know that there is another zero at #(6,0)#, but since this one "bounces off" the x-axis, we know it has an even multiplicity. To keep the polynomial as simple as possible, we will assume the point at #(6,0)# has multiplicity 2.

So far, our polynomial function looks like this: #f(x)=a(x-sqrt5)(x+sqrt5)(x-6)^2=a(x^2-5)(x-6)^2#. To solve for the dilation factor, simply substitute the point #(2,-14)# into the equation to get #-14=a(2^2-5)(2-6)^2#
#-14=-16a->a=7/8#
#:.f(x)=7/8(x^2-5)(x-6)^2#

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