Use the x-intercepts to find the intervals on which the graph of f is above and below the x-axis.
-$$f ( x ) = ( x + 15 ) ^ { 2 }$$
A) above the $x$-axis: $( - \infty , - 15 )$
below the $x$-axis: $( - 15 , \infty )$

B) above the $x$-axis: $( - \infty , - 15 ) , ( - 15 , \infty )$
below the $x$-axis: no intervals

C) above the $x$-axis: $( - 15 , \infty )$
below the $x$-axis: $( - \infty , - 15 )$

D) above the $x$-axis: no intervals
below the $x$-axis: $( - \infty , - 15 ) , ( - 15 , \infty )$

Question

Use the x-intercepts to find the intervals on which the graph of f is above and below the x-axis.
-$$f ( x ) = ( x + 15 ) ^ { 2 }$$
A) above the $x$-axis: $( - \infty , - 15 )$
below the $x$-axis: $( - 15 , \infty )$

B) above the $x$-axis: $( - \infty , - 15 ) , ( - 15 , \infty )$
below the $x$-axis: no intervals

C) above the $x$-axis: $( - 15 , \infty )$
below the $x$-axis: $( - \infty , - 15 )$

D) above the $x$-axis: no intervals
below the $x$-axis: $( - \infty , - 15 ) , ( - 15 , \infty )$

Quizplus · Accepted Answer

The function $f(x) = (x + 15)^2$ is a square of a linear expression, meaning it is always non-negative. It only equals zero when $x = -15$, which is its only x-intercept. Therefore, the graph is above the x-axis for all other values of $x$, and there are no intervals where the graph is below the x-axis.

Use the X-Intercepts to Find the Intervals on Which the Graph