Families of Functions
Families of Lines The family y = mx + b , with m fixed and b varying. m = 1, b = – 2, – 1, 1, 2, 3 m = – 0.2, b = – 2, – 1, 1, 2, 3
Families of Lines The family y = mx + b , with b fixed and m varying m = –2, – 1, 1, 2, 3 b = 1 m = –2, – 1, 1, 2, 3 b = – 2
The Family y = x n The family y = x n , n = 1, 2, 3, … n = 1, 3, 5 n = 2, 4, 6
The Family y = x n n = 101 n = 100
The Family y = x n The family y = x n , n = –1, –2, –3, … n = – 1, – 3, – 5 n = –2, –4, –6
The Family y = x n n = – 100 n = – 101
The Family y = n x n x The family , n = 1, 2, 3, … n = 1, 3, 5 n = 2, 4, 6
The Family y = n x n = 100 n = 101
Polynomials 2 3 3 4 2 4 5 + − + − + + − + 4 x 3 x x 3 x 2 x x 1 x 4 x x
Rational Functions 2 + 1 x 2 3 + 2 x x
Rational Functions 3 − 2 3 x 4 + x 7 x
Rational Functions 8 5 + 3 x
Algebraic Functions + 2 − 2 (2 x ) x 1
Algebraic Functions + − + 2 3 2/3 (2 x ) (1 x x )
A Quick Review of Trigonometry c a θ b θ sin( ) a tan( ) a cos( ) b sin( ) θ = θ = θ = = θ c b c cos( ) θ csc( ) c sec( ) c cot( ) b 1 1 cos( ) θ = θ = = θ = = = θ θ θ a b a cos( ) sin( ) sin( )
z c a x θ θ y b These triangles are similar (all angles equal), so the sides are proportional, that is: a b c h = = ≡ x y z Thus the trig .functions are all equal. hc ha θ hb
ht y θ θ len x ht len ht y y = = x len x If a six ft pole gives a 5 ft shadow, and a building gives a shadow of 250 ft, then the height of the building is 6 = 250 300 5
Trigonometric Identities 2 2 2 = + c c a b a θ b 2 2 + a b θ + θ = = 2 2 sin( ) cos( ) 1 2 c θ Divide both sides by 2 to obtain the identity: cos( ) θ + = θ 2 2 tan( ) 1 sec( ) − = − =− cos( x ) cos( ) x sin( x ) sin( ) x
Two important triangles. π π 4 2 2 3 1 1 π π 6 4 1 3 π π 1 π π = = 1 sin cos = = sin cos 4 4 2 6 3 2 π π 3 = = sin cos 3 6 2
P (cos( θ ), sin( θ )) θ (– 1,0) (1,0) cos( θ ) and sin( θ ) are the projections onto the x and y axes respectively, as the point P goes around the unit circle.
Graphs of the Trigonometric Functions cos( x ) sin( x ) π sin( x ) is equal to cos( x ) shifted to the right by 2 That is: π π = − =− − cos( ) sin x x sin x 2 2
π θ − c a 2 θ b π θ π θ − tan( θ ) = cot 2 − sec( θ ) = csc 2 π θ π θ − − cot( θ ) = tan 2 csc( θ ) = sec 2
tan( x ) cot( x )
sec( x ) csx( x )
sin( ax + b ) for a = 2, b = 1 and a = 6, b = 3
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