Question

18. Find the equation(s) of each of the following graphs. State the domain and range. 6 4 7 2 6 (-3,0) 2 10 8 6 4 2 5 (0) x 2 4 6 8 10 109 90 80 70 60 50 0 30 20 10 12 4 6 8 -10 1 0.5, -12272 -4 -2 -10 4 6 8 10 12 7 6 5 3 2 1 8.716-5-4-313, 133 4 5 6 7 8 با بق ا ال و

Answer 1

of graph (a) The first graph is vertex (0.5 - 12.25). + 12.25 parabola with General equation of such parabola which opens upward is y + k (X-0.5)2 also graph passes through (-3,0) (4,0) which gives - 12.25 +2.0.52 Now, this and K=1. y = this graph is (-9,00 is (-12.25,0) this Domain of Range graph

х -5 y=-12.25 + (x – 0.5)2 2 (-3,0) (4,0) -15 - 10 -5 0 5 10 15 10 (0.5, -12.25) -15

like exponential graph which passes (b) It is looking the y= 1.oka of through (0,1). hence its equation should We take k 07 because we that at a=3 8. Domain graph ich (s, & of range the graph is (0,0). Graph egh will yn hile so see 0.72 be

x »= €0.7x -8 -6 4 -6 -5 -2 2 3 4 5 6

we are that at n=3, ر2 تلا = 25 y = 12.5 one x=0, y = 100, at a= seeing tog (C) In this graph, at y= at y= It means that uvere screasing n, by that half value yat It means radioactive graph in time radioactive substance is loo. will be equati on of graph We observe unit is decreasing to its which half and initial value of a is one y = to ott where =100 Yo constant = ln (2) k = radioactive decay th half time rence K In (2) Equation of geraph y = 100 mm (1) range is (0,100]. domain is [o, oo and

100 Х y=100e-rin 2 -80 -60 -40 20 0 2 10 12

d) In this we are the seeing about is (3 2 graph fluctuation y=4. 4. while amplitude 3 let the equation is of graph 4-3 cos (KC) y = {because at x = 0, y = 1, -sign has been taken in get yo at xo} see that there are peaks of above equation to get 1 we Now, two of oscillations between of a 45. Hence, we take k=2 to such scenerio. Hence can of graph y = 4 - 3 cos (22) (-os, o is [1,1] ich domain Range

ч Х -8- у = 4 – 3 cos(2х) (-3л/2, 7) (-1/2, 7) (п/2, 7) (3л/2, 7) АААА (-л, 1) (0, 1) (л, 1) 0 2 6 -24

we see are that oscillations position x=2 and Again there about mean Amplitude = (5-4-1) 3 2 2+ Besky) yo & si has sign because at n=o, that d=2 see we Therefore, eqr of graph is been taken 4-5.) Now, y = 2 + 3 cos (2K) - Cosak = -1 cos T Cos 2k = ņ -> 2K - T k = T 2 n is 2+3 cos en of graph y = (en) domain is (-2, 0) is (-1,5) Range

(-8,5) (-4,5) (0,5) + y= 2 + 3 cos मत (4,5) (8,5) -8 -2. 0 6 8 10 (-6, -1) (-2,-1) (2, -1) (6, -1) -2