Trig Functions
y=3cos1/4(x-50)
state the amplitude
A = 3
what is the vertical stretch
g(x) = 2f(x) - 5
vertical stretch bafo 2
8x3 x 7x4
=
56x7
(3n-4)2
=
9n2-24n+16
2x2-22x+48=0
what are the two x values?
x1 = 8
x2 = 3
y=3cos1/4(x-50)
state the period
360 / 1/4 = 1440
what is the vertical shift
g(x) = 3f(x) - 8
vertical shift down 8
68x10 / 17x
=
4x9
3m2 + 3m -6
= 3(m2 + m-2)
= 3(m2 + 2m - m -2)
= 3(m+2) x (m-1)
express the following in simplest form
√40
= 2√10
how do you find the amplitude and midline when only given a max and min value
and to find the midline you do | (max + min) /2
{0. 9), (2, 7), (3, 5), (4, 3)}
is the information provided a function or not
yes
(2x2y)6 x 3x9
=
192x13y15
9a2 - 100
=
(3a + 10) (3a - 10)
express the following in simplest form
(2+4√3)(2-4√3)
= - 44
y = -sin(4x + 28) + 1
state the range, how do you find it
range = 0 ≤ y ≤ 2
to find this you need to use the amplitude and midline numbers that they give you and find the max and min
a = 1 c = 1
list the asymptotes for the following equation
h(x) = 1/(x-4) + 2
Y = 2
x = 4
(3/8)-2
=
64/9
25x2 + 25x + 100
= 25(x2 + x + 4)
will the following function have a max or min value?
f(x) = 3(x − 1)(x + 5)
= Minimum value
tan2(x) - sin2(x) = tan2(x)sin2(x)
=
change rs
=[1-cos2(x)]tan2x
=tan2x - cos2x(tan2x)
=tan2(x) - sin2(x) Ls = Rs
a function reflected in the x axis, vertical stretch bafo 3, horizontal compression bafo 1/3, horizontal translation 9 units right, vertical shift 4 units down
determine the equation^
y = -3 √3(x-9) - 4
3-5/ (3-4 + 3-6)
=
3/10
9x2 + 48x + 60
= 3(3x2 + 16x + 20)
= 3(3x2 + 10x + 6x + 20)
= 3(3x+10)+(x+2)
Calculate the value of k such that kx2 − 4x + k = 0 has one root.
K = +- 2