Use the given function to complete parts a) through e) below \( f(x)=-3(x-3)^{2}\left(x^{2}-4\right) \) At which zero(s) does the graph of the function touch the \( x \)-axis and turn around? Select the correct choice below and, if necessary, fill in the answer box to complete your choice A. The graph touches the \( x \)-axis and turns around at the zero(s) (Type an integer or a decimal Use a comma to separate answers as needed \( ) \) B. There are no zeros at which the graph touches the \( x \)-axis and turns around c) Find the \( y \)-intercept by computing \( f(0) \) The \( y \)-intercept is \( \square \) d) Determine the symmetry of the graph Even, \( y \)-axis symmetry Odd, origin symmetry neither \( f(x) \) nor \( -f(x) \)
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The Deep Dive
To find the zero(s) where the graph touches the \( x \)-axis and turns around, we analyze the function: \( f(x)=-3(x-3)^{2}\left(x^{2}-4\right) \). Here, the term \( (x-3)^2 \) indicates that the graph touches the \( x \)-axis at \( x=3 \) because it is squared (denoting a turning point). The factor \( (x^2 - 4) \) gives the additional zeros at \( x = -2 \) and \( x = 2 \), where the graph crosses the \( x \)-axis. Thus, the graph touches the \( x \)-axis at \( x = 3 \). The \( y \)-intercept is found by substituting \( x = 0 \) into the function: \[ f(0) = -3(0-3)^2(0^2-4) = -3(9)(-4) = 108 \] So, the \( y \)-intercept is \( 108 \). To determine the symmetry of the graph, we check the function: 1. For even symmetry, we need \( f(-x) = f(x) \). 2. For odd symmetry, we need \( f(-x) = -f(x) \). After testing, neither holds true for this function. Therefore, the function has neither even nor odd symmetry.