Assuming the function is
f(x) = x^4 + 3x^3-28x^2
Factorising, we get
f(x) = x^2(x+7)(x-4)
Therefore, there is a double root at x = 0 and singular roots at x = -7 and 4
double roots tend to bounce, kind of like the standard quadratic y = x^2
singular roots tend to cross, kind of like a straight line on a graph, y = mx + b
Therefore, it bounces at x = 0, and crosses at x = -7 and 4
In general, a root raised to an even power bounces, and a root raised to an odd power crosses.
No.
Replace the x's with -5 and you get -25=-27 which is not a solution
Answer: 2.5
Step-by-step explanation:
y = rx
substitute numbers in and solve for r
if
y = 10
x = 4
then
10 = r4
r = 10/4 = 2.5
this works for all the other numbers and always gives r = 2.5
Answer:
y = 4/7x
Step-by-step explanation:
(0,0)
(7,4)
up 4 right 7
4/7
y = 4/7x
Answer:
Step-by-step explanation: