Basic Functions
Find the vertex and the x-intercept(s) of the function f(x) = 25 - x^2
Vertex: (0, 25) x-intercepts: (+/-5, 0)
The domain of (x+1)1/2 in interval notation
[-1, inf)
The leading term in f(x)=(x+1)3(5x-1)(x-3)4
5x8
The solution to 4x=7
x=log47 because log and exponential functions of same base are inverses
The conversion of 160 degrees to radians
160/x = 180/pi
x=160pi/180
=8pi/9
The equation of a line in point-slope form that passes through the points (1,2) and (-3,4)
m=2/-4 = -1/2
y-2=-1/2 (x-1)
The domain of (2x+1) / (x-5)(2x) in interval notation
x can't be zero or 5 but can be everything else
(-inf, 0) U (0,5) U (5, inf)
The end behavior of f(x)=-3x7+5x4-4x3+9
Negative LC, odd degree so end behavior is like -x3
53x-1=52x
3x-1=2x
x=-1
(or can take log5 on both sides)
The amplitude and period of y=2sin(4x)
amplitude=2
frequency=4, so period=2pi/4 = pi/2
The composition f(g(h((x))) if f(x)=3x2-4, g(x)=2x+1, h(x)=1/x
3(2/x+1)2-4
The range of y=5x-3+1 and the equation(s) of any asymptote(s)
(1, inf) and horizontal asymptote is y=1
The result of 2x3-6x+1 divided by x-3
2x2+6x+12+ 37/(x-3)
The value of log3(815)
log3((92)5)
log3(((32)2)5)
log3(320)
=20
The value of tan-1(-1)
The range for tan-1(x) is (-pi/2, pi/2)
tan-1(-1) means we need x so that sin x / cos x = -1, which means sine and cosine must be the same value at this angle, but opposite signs (Q4)
tan-1(-1) = -pi/4
The solution to |-2x+1|<5 in interval notation
-2x+1<5 and -2x+1>-5
-2x<4 and -2x>-6
x>-2 and x<3
(-2,3)
The domain, asymptote(s), and x-intercept of y=log5(2x+3)-1
Domain: 2x+3>0, so x>-3/2
VA: x=-3/2
x-int: (1,0) because 0=log5(2x+3)-1, so 1=log5(2x+3), so 51=2x+3, which gives x=1
The hole(s) of the rational function
f(x)=(x+1)(x-1) / (3x+2)(x-1)
x=1 and y=(1+1)/(3(1)+2)=2/5
(1,2/5)
The solution to log7(x-2)+log7(x+3)=log714
x=4
From a point on the ground 47 feet from the foot of a tree, the angle of elevation to the top of the tree is 35º. Find the height of the tree to the nearest foot. (Calculator allowed)
33
The completed square process to convert the quadratic function f(x) = x^2 - 2x + 5 into standard/vertex form
x^2-2x
(x-1)(x-1) = x^2-2x+1
(x^2-2x+5)+1-1
(x^2-2x+1)+4
(x-1)^2+4
The domain and range of y=2 sin(2x-pi)-3
D: all real numbers
R: [-5,-1]
The asymptote(s) and the x-coordinate of all holes of the rational function
f(x)=(x+1)(2x-1) / (3x+2)(2x-1)(-x+3)
VA: x=-2/3 and x=-3
HA: y=0
Hole: occurs at x=1/2
The solution of log2(x+1)-log2(x-4)=3
x=33/7
The solution to 2tan(x)sin(x)+2sin(x)=tan(x)+1 on [0,2pi]
pi/4, 5pi/4, pi/6, 5pi/6