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Algebra 15 Online
OpenStudy (anonymous):

Who Wants To Be My Partner ?? *fan and medal awarded*

OpenStudy (anonymous):

my assignment is collaborative. i dont have a partner. does anyone want to be my partner and help me finish ?

OpenStudy (anonymous):

y = ax^3 + bx^2 + cx + d Graph this using any graphing technology of your choice. However, replace the variables a, b, c, and d with numbers. For example, you could graph y = 2x^3 + 3x^2 – x + 2. Observe the graphed outcomes together (or independently if working alone).

OpenStudy (anonymous):

haha ight just to let you kno im ranked 3 at pearland high

OpenStudy (anonymous):

good for u

OpenStudy (anonymous):

first dy/dx= 3ax² +bx + c

OpenStudy (anonymous):

1 Sec

OpenStudy (anonymous):

k

OpenStudy (anonymous):

Ok so.

OpenStudy (anonymous):

so 3a(-2)² +b(-2) + c = 0

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

and 3a(2)² +b(2) + c = 0

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

Now if you substitute -2 in for x and 6 in for y you get another linear equation. 6 = a(-2)³+b(-2)²+c(-2)+d

OpenStudy (anonymous):

okay i think u misunderstood.

OpenStudy (anonymous):

it doesnt need to be explained to me. its a collaborative assignment. this part is what im supposed to have my partner do. im not supposed to do it. i already done my part.

OpenStudy (anonymous):

Oh OK i get you now.

OpenStudy (anonymous):

so can u fulfill the partner status ?

OpenStudy (anonymous):

Well i would but i cant; i'm currently making a program for a company. An Anti Virus system and iv'e been busy with that. If you want fan me and i will msg you if i have time on my hands?

OpenStudy (anonymous):

well if u have the time to explain it im sure u have the time to jus graph it.

OpenStudy (anonymous):

i jus need y = 2x^3 + 3x^2 – x + 2 put onto a graph and i can turn it in

OpenStudy (anonymous):

Why do you dislike me? I just don't have time to do it.

OpenStudy (anonymous):

if u have time to explain it, then u have time to graph it. it would actually take less time to graph it.

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