Keepin’ It Imaginary
SOLVE: i3
−i
Find the roots: x2 = 36
x=±6
Find the vertex: f(x)=x2+4x−1
(−2,−5)
solve the radical: √6x+1=4
x=5/2
Solve the following.
|2x+1|-6=19
x=12, −13
−√−81
−9i
Factor by completing the square:
x2+ 5x − 14 = 0
x=−7,2
Find the vertex, AOS, & domain/ranges: f(x)=x2−4x+3
Vertex: (2,−1)
AOS: x=2
Domain: (−∞,∞)
Range: [−1,∞)
Solve for x:
x+3/ x-2= 4/ (x+1) - 4/ x2-x-2
x=±i√ 15
In absolute value inequalities, this word describes the solution where a number must satisfy both conditions at the same time.
What is AND?
(8 + 2i) − (7 − 4i)
1+6i
2x2−5x−12=0
x=4, -3/2
Find vertex, AOS, inc/dec, and dom/range: f(x)=−2x2+8x−3
vertex: (2,5)
AOS: x=2
inc/dec: (−∞,2) to (2,∞)
dom/range: (−∞,∞) , (−∞,5]
Solve for x.
(x+4)/4 −(2x−7)/3 = 0
x=8
|2x−5|<7
−1<x<6
(13+2i)(11−4i)
151−30i
Solve.
f(x)=x2−4x−24
x= 2±2√7
Find the vertex, AOS, max/min, dom/range, and inc/dec for the function: f(x)=−3(x−2)2+5
Vertex: (2,5)
Axis: x=2
Maximum: 5
Domain: (−∞,∞)
Range: (−∞,5]
Increasing: (−∞,2)
Decreasing: (2,∞)
Solve for x.
1/x − 1/ x+1 +1/ x+2 = 0
x=-1±i
5−3|4x−7|=−10
x=3 OR x=1/2
7−2i/3+4i
13/25 + 34/25i
3x2−10x+14=0
5/3 ± √17/3i
A quadratic function G(x) has a vertex at (4,−2) and passes through the point (−1,53). Determine the final equation in vertex form.
G(x)=11/5 (x−4)2−2
Solve.
(x-2)/(x+4) =6 / (x-1) - 6 / x2+3x-4
x= 9 ± √ 145 / 2
2−3|5x−4|=−7
x=1/5 OR x=7/5