Find the x-intercepts of the polynomial function. State whether the graph crosses the x-axis, or touches the x-axis and
turns around, at each intercept.
-$$f ( x ) = - x ^ { 2 } ( x + 4 ) \left( x ^ { 2 } + 1 \right)$$
A) 0 , touches the $x$-axis and turns around;
4 , crosses the $x$-axis
B) 0 , touches the $x$-axis and turns around;
$- 4$, crosses the $x$-axis;
$- 1$, crosses the x-axis;
1, crosses the x-axis;
C) 0 , touches the $x$-axis and turns around;
$- 4$, crosses the $x$-axis
D) 0 , touches the $x$-axis and turns around;
$- 4$, crosses the $x$-axis;
$- 1$, touches the $x$-axis and turns around

Question

Find the x-intercepts of the polynomial function. State whether the graph crosses the x-axis, or touches the x-axis and
turns around, at each intercept.
-$$f ( x ) = - x ^ { 2 } ( x + 4 ) \left( x ^ { 2 } + 1 \right)$$
A) 0 , touches the $x$-axis and turns around;
4 , crosses the $x$-axis
B) 0 , touches the $x$-axis and turns around; 
$- 4$, crosses the $x$-axis;
$- 1$, crosses the x-axis;
1, crosses the x-axis;
C) 0 , touches the $x$-axis and turns around;
$- 4$, crosses the $x$-axis
D) 0 , touches the $x$-axis and turns around; 
 $- 4$, crosses the $x$-axis;
$- 1$, touches the $x$-axis and turns around

Quizplus · Accepted Answer

The x-intercepts are found by setting $f(x) = 0$, which gives $x = 0$ and $x = -4$. The term $x^2$ indicates that at $x = 0$, the graph touches the x-axis and turns around because the power is even. The term $(x + 4)$ indicates that at $x = -4$, the graph crosses the x-axis because the power is odd. The term $(x^2 + 1)$ does not affect the x-intercepts because it never equals zero for any real value of $x$.

Find the X-Intercepts of the Polynomial Function f(x)=−x2(x+4)(x2+1)f ( x ) = - x ^ { 2 } ( x + 4 ) \left( x ^ { 2 } + 1 \right)f(x)=−x2(x+4)(x2+1)

Find the X-Intercepts of the Polynomial Function $f ( x ) = - x ^ { 2 } ( x + 4 ) \left( x ^ { 2 } + 1 \right)$