Question

precalc part two @shadow

Answer 1

@dude

Answer 2

well at least post it

Answer 3

How would you graph -2x^2(x-3)^3(x+2)^2

Answer 4

by graphing it duh

Answer 5

Jokes, it would help to find the roots of the equation

Answer 6

(x-intercepts)

Answer 7

zero, 3, -2
bounce on zero and neg two

Answer 8

\(\color{#0cbb34}{\text{Originally Posted by}}\) @dude
by graphing it duh
\(\color{#0cbb34}{\text{End of Quote}}\)
._.

Answer 9

Good, you're still missing the root for -2x^2 though

Answer 10

zero?

Answer 11

Ya

Answer 12

\(\color{#0cbb34}{\text{Originally Posted by}}\) @lowkey
zero, 3, -2
bounce on zero and neg two
\(\color{#0cbb34}{\text{End of Quote}}\)
Z E R O

Answer 13

LMOAOOOOAOO

Answer 14

How would you find the end behaviors?

Answer 15

Usually to have end behaviors your equation has to be some sort of funky (rational functions for example)

In this case, they would go on forever

Answer 16

Alright, thanks dude <3

Answer 17

End behaviors for polynomial functions:
You can see from the graph that the function approaches infinity (function is growing large positive without bound) as \(x \to -\infty\) and the function approaches negative infinity (function is growing large negative without bound) as \(x \to \infty\)

For a more general algebraic sense, we see that \(y=-2x^2(x-3)^3(x+2)^2\) is an odd polynomial function as the multiplicities of the roots (exponents on the roots) sum to an odd number (\(2 + 3 + 2 = 7\)); the end behavior therefore goes in different directions. The negative coefficient indicates the following end behavior:
As \(x \to -\infty\), \(y \to \infty\) and as \(x \to \infty\), \(y \to -\infty\).
(For \(y = mx+b\) with negative slope, it also follows this end-behavior description.)

Answer 18

Hello smartie would you like fifty dollars to cheat off my trig tests pls and ty