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, ∞)
The degree of the function f(x)=(x+1)3(5x-1)(x-3)4
The solution to
4x = 8
x= 3/2
160 degrees in radians
8π/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
(-∞, 0) U (0,5) U (5, ∞)
The end behavior of f(x)=-3x7+5x4-4x3+9
Negative leading coefficient, odd degree so end behavior is like -x3, starts high on the left, ends low on the right.
The solution to
53x-1=25x
53x-1=52x
3x-1=2x
x = 1
The amplitude of y=2sin(4x)
amplitude=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
(1, ∞)
The result of
2x2-x-15 divided by x-3
2x+5
The value of log3(815)
log3((92)5)
log3(((32)2)5)
log3(320)
=20
sin(5π/3)
-√3/2
The solution to |-2x+1|=5
x = -2 and x = 3
The domain of y=log5(2x+3)-1
Domain: 2x+3>0, so x>-3/2 or (-3/2, ∞)
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)=0
x = [-1+sqrt(29)]/2
From a point on the ground 17 feet from the foot of a tree, the angle of elevation to the top of the tree is 60º. Find the height of the tree.
17√3 feet
The quadratic function f(x) = x^2 - 2x + 5 converted into vertex form
(x-1)^2+4
The domain and range of y=2 sin(x)-3
D: all real numbers
R: [-5,-1]
The asymptote(s) of the rational function
f(x)=[(x+1)(2x-1)] / [(2x-1)(-x+3)]
VA: x = 3
HA: y = -1
The solution of log2(x+1)-log2(x-4)=3
x=33/7
The value of
tan-1(-1)
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, and keeping the domain in mind, can only be in (Q4)
tan-1(-1) = -π/4