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
1.404
Write a polynomial in factored form with zeros at 1,-1 and 2.
(x+1)(x-1)(x-2)
Write in vertex form (complete the square):
x2-2x+5
(x-21)2+4
The domain of (2x+1) / (x-5)(2x)
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
The solution to 53x-1=25
1
Write the polynomial that has zeros at -2 (with multiplicity of 3) and a zero at 4.
(x+2)3(x-4)
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(2x-1)=4
41
Write a polynomial with zeros at i and 3.
(x2+1)(x-3)
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
Factor completely and find all zeros: x3+3x2-4x-12
(x+2)(x-2)(x+3)
2,-2,-3
Write in vertex form (complete the square):
3x2-12x+16
3(x-1)2+4
The domain of y=sqrt(3x-1)+2
D: x >= 1/3
R: y>=2
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
Factor completely and find all zeros:
2x4-9x3+9x2+x-3
(x-1)2(x-3)(2x+1)