Question

find the zeros of the function, and then describe the behavior of the graph at each zero.
$x^3 - 12x^2 + 35x$

the zero(s) is/are 0,5,7.
(use a comma to separate answers as needed.)
describe the behavior of the graph at each zero. select the correct choice below and fill in the answer box(es) to complete your choice.
(use a comma to separate answers as needed.)
a. the graph crosses the x - axis at \\(\square\\). the graph touches the x - axis without crossing it at \\(\square\\).
b. the graph crosses the x - axis at \\(\square\\). the graph never touches the x - axis without crossing it.
c. the graph never crosses the x - axis. the graph touches the x - axis without crossing it at \\(\square\\).

Explanation:

Step1: Factor the function

First, factor out the common factor \( x \) from the polynomial \( x^3 - 12x^2 + 35x \). We get \( x(x^2 - 12x + 35) \). Then, factor the quadratic \( x^2 - 12x + 35 \). We find two numbers that multiply to 35 and add to -12, which are -5 and -7. So, \( x^2 - 12x + 35=(x - 5)(x - 7) \). Thus, the factored form of the function is \( f(x)=x(x - 5)(x - 7) \).

Step2: Analyze the multiplicity of each zero

For a zero \( r \) of a polynomial function, if the multiplicity of \( r \) is odd, the graph crosses the x - axis at \( r \); if the multiplicity is even, the graph touches the x - axis (and turns around) at \( r \). In the factored form \( f(x)=x(x - 5)(x - 7) \), each factor \( x \), \( (x - 5) \), and \( (x - 7) \) has a multiplicity of 1 (since the exponent of each linear factor is 1). Since 1 is an odd number, the graph will cross the x - axis at each of the zeros \( x = 0 \), \( x=5 \), and \( x = 7 \).

Answer:

The correct choice is B. The graph crosses the x - axis at \( 0,5,7 \). The graph never touches the x - axis without crossing it.