Question

1. Given a function. y = (x - 493 a. State the domain. b. Find the x-intercept and the y-intercept. C. Is there any symmetric? Find asymptotes, if any. d. Find y' and critical value.

Answer 1

f(0) = (2-4) a) The domain of a function is the set of input or argument values for which the function is real and defined. " The function above has no undefined points nor domain constraints. Therefore, the domain is o Laco b) x-intercept is a point on the graph where y=0 Solve (x-4)2/=0 .. x-4=0 x=4 x-intercepts (4,0) y-intercept is a point on the graph where x=0 y = (0-4)943 - y = 1-4)% = (-13% (433 213 - y = 42 2 C 213 V-intercept 10,443) D A graph is symmetric about the al-axis if . every you the graph can be replaced with y -ya (x -403213 y= -(x-43213 1. → –(x-4)243 + (0-4)." . No symmetry about the al-axis 2) A graph is symmetric about the goaxis if every x on the graph can be replaced with a

2212 y=[(-x) -432/3 . y= (-x-4) 2/3 . (--4)23 (0-4)3 No symmetry about the y-axis 3) A graph is symmetric about the origin if every ac on the graph can be replaced with i and every on the graph can be replaced with-y -y=[(-x)-4723 ye-(-0–423 = -(-22-4)23 € (0-203743 No symmetry about the origin. Therefore the equation has no Symmetry. -> Asymptotes 1) vertical Asymptotes Go over every undefined point aza and check if at least one of the following is true: lim f(x) = to and lim f(x) = to 2797 The function f(x) has no undefined pointes, Therefore No vertical Asymptotes 2). Horizontal Asymtotes Check if at a to the function behave as a line, y=matb * Compute lim f(x) to find m e => lim ((-1)2) = 0 m=0] X+- 01 Compute lim f(x) - mx to find b ata 1 X- no ol -00

at-o 15 lim (11-48²_0.x) => Directed Infinity [ -1)"}] The result is not te constant, Therefore No horizontal Asymptotes * Compute lim f(x) to find m =) lim (1-4)² =0 [m zo] x 700 x Compute lim f(x)-mx to findb Xto xo =) lim ((x-4)²3_0.x) => as ato The result is not a finite constant, Therefore No horizontal Asymptotes a) f(x) => y = (2-4) 2/3 (X=43743 Chain aule dif(u= of die Let u=d-4 and f = 02/ - By Chain rule dy => y = 2 3 4 (X-4) y' = 2 U23-1) [10]

y' = 2 / Ull substitute back u= (2-4) .ly' = 2/3 (0-4) 13 find critical point put y'=0 2 (0-4) =0 To Ol -4 = [2=4] E [-00,oo critical point 1 x=4 is a