Since we are given that the polynomial has x-intercepts or zeroes at (+-sqrt5,0)(±√5,0), we know that the polynomial must have factors (x-sqrt5)(x−√5) and (x+sqrt5)(x+√5). In addition, we know that there is another zero at (6,0)(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)(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)^2f(x)=a(x−√5)(x+√5)(x−6)2=a(x2−5)(x−6)2. To solve for the dilation factor, simply substitute the point (2,-14)(2,−14) into the equation to get -14=a(2^2-5)(2-6)^2−14=a(22−5)(2−6)2
-14=-16a->a=7/8−14=−16a→a=78
:.f(x)=7/8(x^2-5)(x-6)^2
