3f(x) has the same zeroes as f(x), since
3(x+4)(x-2) = 0
is zero at the same places where f(x) is.
f(-x/2) = (-x/2 + 4)(-x/2 - 2)
= 1/2(-x+8) * 1/2 (-x - 4)
= 1/4 (x-8)(x+4)
so it has zeroes at 8 and -4
Note that using the original zeroes of -4 and 2,
-1/2(8) = -4
-1/2(-4) = 2
so f(ax) has zeroes at x1/a and x2/a
Use transformations and the zeros of the quadratic formula f(x)=(x+4)(x-2) to determine the zeroes of each of the following functions:
a. y=3f(x)
f. y=f(-1/2x)
I have no idea how to start this. Any help appreciated. Thanks.
1 answer
1 answer