Linear & Quadratics
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 inverse of f(x)=4x-1.
Swap variables, x=4y-1 and solve for new y:
x+1=4y
(x+1)/4=y=f-1(x)
2|4-x|=6
|4-x|=3
4-x=3 4-x=-3
x=1, 7
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
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
State all transformations that occurred to y=2|3x-12|+1 using the parent function y=|x|
y=2|3(x-4)|+1
Vertical stretch by a factor of 2, horizontal compression by a factor of 1/3, right 4, up 1
The quadratic function f(x) = x2 - 4x + 5 in standard/vertex form
h=-(-4)/2(1)=2
k=f(2)=22-4(2)+5=1
y=(x-2)2+1
The domain and range of y=-(3x+2)1/2+1
D: [-2/3, inf)
R: The outputs (y values) are effected by the negative on the outside, which reflects the square root function over the x axis, and by the +1 on the outside, which shifts it up one unit. Thus the range is (-inf, 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)
Restrict the domain of the function y=(x-1)2+3 so that it has an inverse that is also a function.
The vertex of this quadratic function is (1,3) and it opens up. To make this a function with an inverse, we need it to be one-to-one, so we have to restrict its domain in order to only have half of a parabola. To accomplish this, we can either make its restricted domain [1,inf) or (-inf, 1].
The solution set 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 vertex and range (in interval notation) of y=2(x+3)2+5.
Range = shifted to the left 3 units and up 5 units, parabola opens up, so the smallest output is y=5, making the range [5,inf)
The domain and range of y=(x-1)/(x2-1) in interval notation.
D: x cannot be 1 or -1, so (-inf, -1)U(-1,1)U(1,inf)
R: The HA is y=0, there is a hole at (1, 1/2), and the VA is x=-1. A rough sketch of the graph shows that we do not cross the HA, so the only y values we never get are y=0 and y=1/2, making the range (-inf, 0) U (0,1/2) U (1/2, inf).
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
Rewrite h(x)=|2x-3|+1 as a composition of two functions, so that h(x)=f(g(x)).
Inner function is g(x)=2x-3
Outer function is f(x)=|x|+1
Using the parent function y=x3, write a new function that transforms this by:
reflection over the y-axis, vertical compression by a factor of 1/2, shift to the right 2 units, and down 5.
y=1/2(-(x-2))3-5
or simplified, y=1/2(-x+2)3-5