Answer:
x- intercepts are where the graph crosses the x-axis i,e
substitute y=f(x)=0 to solve for x;
Given the function:
A)
[tex]f(x) = \log x - 1[/tex]
By the definition of x-intercepts;
Substitute y=0 and solve for x;
0 = log x -1
Add 1 to both sides of an equation;
0+1 = log x -1 +1
Simplify:
1 = log x
[tex]10^1 = x[/tex] [Using [tex]\log_{10} x = b[/tex] ⇒[tex]x =10^b[/tex] ]
Simplify:
x =10
Therefore, x-intercepts is ( 10, 0)
B)
[tex]f(x) =-(\log x - 2)[/tex]
By the definition of x-intercepts;
Substitute y=0 and solve for x;
0 = -(log x - 2) or
log x -2 = 0
Add 2 to both sides of an equation;
log x -2 +2 = 0+2
Simplify:
log x = 2
[tex]x = 10^2[/tex] [Using [tex]\log_{10} x = b[/tex] ⇒[tex]x =10^b[/tex] ]
Simplify:
x =100
Therefore, x-intercepts is ( 100, 0)
C)
[tex]f(x) =\log (-x -2) [/tex]
By the definition of x-intercepts;
Substitute y=0 and solve for x;
0 = log (-x - 2) or
log (-x -2) = 0
[tex]-x-2= 10^0[/tex] [Using [tex]\log_{10} x = b[/tex] ⇒[tex]x =10^b[/tex] ]
Simplify:
-x -2 =1 or
x+2 = -1
subtract 2 from both sides we get
x+2 -2 = -1 -2
Simplify:
x = -3
Therefore, x-intercepts is (-3, 0)
D)
[tex]f(x) =-\log -(x -1) [/tex]
By the definition of x-intercepts;
Substitute y=0 and solve for x;
0 = -log -(x - 1) or
log -(x -1) = 0
[tex]-(x-1)= 10^0[/tex] [Using [tex]\log_{10} x = b[/tex] ⇒[tex]x =10^b[/tex] ]
Simplify:
-(x -1) =1 or
x - 1 = -1
Add 1 to both sides, we get,
x -1 +1 = -1 +1
Simplify:
x = 0
Therefore, x-intercepts is (0, 0)
Match the function to their intercepts:
(0,0) <-------> __D___
(-3,0) <------> __C___
(10,0) <------> __A___
(100,0) <----> __B___
