+ 4 − 21 = −21 x2 x The zeros of are −7 and 3. Likewise, the f + 4 = x2 x of are −7 and 3. x-intercepts f 0 ( ) ( = −21 c. + = f x 4) x x 0 ( = 1( − (−5))( − 3) f x) x x = 0 or + 4 = 0 x x = ( + 5)( − 3) = + 2x x x x2 = −4 x −15 For = 2 : a ( = 2( − (−5))( f x) x x The solutions ( ) = −21 −4 are and 0. − f x 3) = 2( + 5)( − x x (−4, (0, −21) −21) Thus, the points and are 3) on the graph of f. = 2( + 2 −15) = 2 + 4 x2 x x2 x − 30
The zeros of f are −7 and 3. Likewise, the x-intercepts of f are −7 and 3. c. f ( x ) = −21 x2 + 4 x − 21 = −21 x2 + 4 x = 0 x ( x + 4) = 0 f ( x) = 1( x − (−5))( x − 3) = ( x + 5)( x − 3) = x2 + 2x −15 x = 0 or x + 4 = 0 x = −4 For a = 2 : f ( x) = 2( x − (−5))( x − 3) The solutions f ( x ) = −21 are −4 and 0. = 2( x + 5)( x − 3) = 2( x2 + 2 x −15) = 2 x2 + 4 x − 30 Thus, the points (−4, −21) and (0, −21) are on the graph of f.
Downloadable-Solution-Manual-for-Precalculus-Concepts-Through-1.docx