Stitz-Zeager_College_Algebra_e-book

Hx 5 4 3x 3x2 x solution 1 in the expression for f

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1)2 + 3(−1) + 4 = −(1) + (−3) + 4 =0 Similarly, f (0) = −(0)2 + 3(0) + 4 = 4, and f (2) = −(2)2 + 3(2) + 4 = −4 + 6 + 4 = 6. 2. To find f (2x), we replace every occurrence of x with the quantity 2x f (2x) = −(2x)2 + 3(2x) + 4 = −(4x2 ) + (6x) + 4 = −4x2 + 6x + 4 The expression 2f (x) means we multiply the expression f (x) by 2 2f (x) = 2 −x2 + 3x + 4 = −2x2 + 6x + 8 46 Relations and Functions Note the difference between the answers for f (2x) and 2f (x). For f (2x), we are multiplying the input by 2; for 2f (x), we are multiplying the output by 2. As we see, we get entirely different results. Also note the practice of using parentheses when substituting one algebraic expression into another; we highly recommend this practice as it will reduce careless errors. 3. To find f (x + 2), we replace every occurrence of x with the quantity x + 2 f (x + 2) = = = = −(x + 2)2 + 3(x + 2) + 4 − x2 + 4x + 4 + (3x + 6) + 4 −x2 − 4x − 4 + 3x + 6 + 4 −x2 − x + 6 To find f (x) + 2, we add 2 to the expression for f (x) f (x) + 2 = −x2 + 3x + 4 +...
View Full Document

Ask a homework question - tutors are online