Question

equations of polynomial help

Answer 1

so far i know the degree, turning point, and its negative

Answer 2

That's a posotive function ik cause I just took a quiz on this stuff

Answer 3

f(x) x(x-2)(x+1)^2

Answer 4

so its positive not negative?

Answer 5

If the leading coefficent is positive and the degree of the polynomial is even, then both ends of the graph point up. We see both sides pointing up, so what can we say?

Answer 6

That it is positive

Answer 7

correct, also @WelpHereWeGo don't give direct answers.

Answer 8

I actually had that answer in the first place just didn’t know if it was right

Answer 9

So now you need help with the zero's and multiplicities?

Answer 10

I got -1,2 for that

Answer 11

okay, anyways the equation that welphere provided is wrong,

Answer 12

Is it x^2(x-2)(x+1)

Answer 13

So to find multiplicities based on the graphs, we have to see the behaviors close to the intercepts, if the function bounces off the x-intercept (doesn't cross x-axis) then that would have an even multiplicity. Let's look at the root at x=2, if we see, close to that general area we can see a parabolic like shape. Do you see a parabolic like shape around x=-2?

Answer 14

yes wouldn’t it be cubed?

Answer 15

So like (x+2)^3

Answer 16

can yall help me

Answer 17

sorry, i was afk

Answer 18

anyways we see a quadratic like shape right?

Answer 19

@nikkixmarie

Answer 20

\(\color{#0cbb34}{\text{Originally Posted by}}\) @darkknight
correct, also @WelpHereWeGo don't give direct answers.
\(\color{#0cbb34}{\text{End of Quote}}\)
😗 ma bad