Houghton Mifflin Harcourt Algebra 1, 2015
HM
Houghton Mifflin Harcourt Algebra 1, 2015 View details
Module 13 Assessment Readiness
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Exercise 1 Page 496

a We are given a piecewise function and are asked to determine whether each of the given points is a solution of f(x). To do so, we will evaluate each x-value and see if the answer matches the given y-value.

f(x)= 3 &ifx<2 - x+1 &if2≤ x <6, x &if x ≥ 6 Is ( - 5, 3) a solution of f(x)? As - 5< 2, we will use the piece f(x)=3. f( - 5)= 3 Since 3= 3, the point (- 5,3) is a solution of f(x).

b Is ( 2, - 1) a solution of f(x)? As 2 is greater than or equal to 2 and less than 6, we will use the piece f(x)=- x+1.

f( 2)=- 2+1 ⇔ f( 2)= - 1 Since - 1= - 1, the point (2,- 1) is a solution of f(x).

c Is ( 8, - 7) a solution of f(x)? As 8≥ 6, we will use the piece f(x)=x.

f( 8)= 8 Since - 7≠ 8, the point (8,- 7) is not a solution of f(x).