+130584 We need to solve the inequality
x2 + x -20 > 0
In these cases, I usualy like to just set this up as an equality and find some "test" points !!!
So we have
x2 + x -20 = 0 and factoring, we have
(x + 5) (x -4) = 0 and setting each factor to 0, we have that x = - 5 and x = 4
This tells us that the possible solutions come from these intervals (-∞, -5), (-5, 4) or (4,∞)...usually, in these quadratic inequalities, if the middle interval "works," the two "outside" intervals don't - or vice-versa !!!
Let's pick a point in the interval (-5, 4) and see if it "works" in the original problem - I always like "0"
So we have
(0)2 + (0) - 20 > 0 and that clearly isn't true
So...my 'guess' is that the solutions come from the intervals (-∞, -5) and (4,∞)
Here's the graph of y = x2 + x - 20
https://www.desmos.com/calculator/owbbudodv9
Ah......just as we suspected !!! ......the function is > 0 on (-∞, -5) and (4,∞) !!!
+130584 If we have
y = 4cosx -1
This is the normal cosine graph with an amplitude of 4 that is shifted "down" one unit. Thus, the normal cosine graph is shifted "down' by one unit and "stretched" vertically.
Here are the graphs of y = cos x and y = 4cosx -1 https://www.desmos.com/calculator/fnvfsqgvgm
Note that the periods of the two functions are exactly the same, so there is no horizontal "stretching."
