Question

The function fff is given in three equivalent forms. Which form most quickly reveals the zeros (or "roots") of the function? Choose 1 answer: Choose 1 answer: (Choice A) A f(x)=-3(x-2)^2+27f(x)=−3(x−2) 2 +27f, (, x, ), equals, minus, 3, (, x, minus, 2, ), squared, plus, 27 (Choice B) B f(x)=-3(x+1)(x-5)f(x)=−3(x+1)(x−5)f, (, x, ), equals, minus, 3, (, x, plus, 1, ), (, x, minus, 5, )(Choice C) C f(x)=-3x^2+12x+15f(x)=−3x 2 +12x+15f, (, x, ), equals, minus, 3, x, squared, plus, 12, x, plus, 15 Write one of the zeros. xxx =

Answer 1

Answer:

(B) $f(x)=-3(x+1)(x-5)$

x=5

Step-by-step explanation:

Given the three equivalent forms of f(x):

$f(x)=-3(x-2)^2+27\\f(x)=-3(x+1)(x-5)\\f(x)=-3x^2+12x+15$

The form which most quickly reveals the zeros (or "roots") of f(x) is

(B) $f(x)=-3(x+1)(x-5)$

This is as a result of the fact that on equating to zero, the roots becomes immediately  evident.

$f(x)=-3(x+1)(x-5)=0\\-3\neq 0\\Therefore:\\x+1=0 or x-5=0\\The zeros are x=-1 or x=5$

Therefore, one of the zeros, x=5