Question

I'm doing "Polynomial and Rational Functions" in my College Algebra Course. I need help doing this problem:
76.) f(x)=-x^4 + 6x^3 - 9x^2.
The instructions say "Sketch the graph of each function."
Anyone willing to solve the problem and explain to me how they did it? Thank you! :)

Answer 1

have you done quadratic factoring yet?

Answer 2

Yes I have

Answer 3

\(\large {
f(x)=-x^4 + 6x^3 - 9x^2\implies
\begin{array}{cccllll}
-x^2(x^2&-6x&+9)\\
&\uparrow &\uparrow \\
&-3-3&-3\cdot -3
\end{array}
}\)

any ideas on the zeros or "solutions"?

Answer 4

Wouldn't you factor it as \[x ^{2}(x ^{2}-4x+3)\] ?

Answer 5

hmmm \((x^2-6x+9)?\) ....

Answer 6

....not sure where the -4x fits in or +3 though

Answer 7

Wait, never mind. I was looking at the wrong problem in my book :D

Answer 8

you're correct

Answer 9

so.... what.... do we end up with anyway?

Answer 10

After factoring, it should be x=0, x=3, and x=3. So, wouldn't the zero multiplicity be 2 and 1?

Answer 11

No, I think the zero multiplicity is actually 2 because after factoring, it would look like this: \[x ^{2}(x-3)^{2}\]

Answer 12

2 , you're correct

Answer 13

hmmm so... we know that \(\large {
f(x)=-x^4 + 6x^3 - 9x^2\implies
\begin{array}{cccllll}
-x^2(x^2&-6x&+9)\\
&\uparrow &\uparrow \\
&-3-3&-3\cdot -3
\end{array}
\\ \quad \\
0=-x^2(x-3)^2\implies
\begin{cases}
x=0&2\ multiplicity\\
x=3&2\ multiplicity
\end{cases} }\)

Answer 14

when the multiplicity is an EVEN number
recall that the graph only makes a U-Turn at the point, it doesn't cross the x-axis

so... to know where is UP or DOWN..... pick one value for "x" before and after those u-turn points and graph it away :)

Answer 15

wouldn't the graph be upside down because the leading coefficient is a negative?

Answer 16

I'd think so, yes

Answer 17

So what would the graph look like?