find a cubic polynomial...

[xomandi]

find a cubic polynomial with equation y=ax^3 + bx whos tangent line at (2,-6) is y= -19x + 32

my work:
so we know that at (2,-6) the slope of the tangent line = -19... therefore, at (2,-6) the derivative of the cubic polynomial = -19.

could i differentiate the cubic polynomial with the variables and set it equal to -19?
y' = -19
y' = 3ax^2 + bx
3ax^2 + bx = -19

then substitute in the x,y values?? how would i solve for a and b?

does this make any sense? please help

 

[galactus]

You're thinking on the right lines.

You can use the two equations to find a and b.

\(\displaystyle \L\\y=ax^{2}+bx\)
\(\displaystyle \L\\y'=3ax^{2}+b\)

Now, use the given (2,-6) and the fact of slope of this line is -19.

\(\displaystyle \L\\-6=a(2)^{3}+2b\)
\(\displaystyle \L\\-19=3a(2)^{2}+b\)

Solve for a and b.

If you do it right, you should get:

 

[xomandi]

oh man duhhh!!! thank you so much!

 

[xomandi]

You're thinking on the right lines.

You can use the two equations to find a and b.

\(\displaystyle \L\\y=ax^{2}+bx\)
\(\displaystyle \L\\y'=3ax^{2}+b\)

Now, use the given (2,-6) and the fact of slope of this line is -19.

\(\displaystyle \L\\-6=a(2)^{3}+2b\)
\(\displaystyle \L\\-19=3a(2)^{2}+b\)

Solve for a and b.

If you do it right, you should get:

i got a = -2 and b = 5

so

y = -2x^3 + 5x

is that right??

 

[galactus]

Did you graph it?. Check it using your data.

Yes, that is right.

 

[xomandi]

yes i did. thanks so much!