# tangent line/derivative

• Oct 23rd 2008, 04:04 PM
algebra2
tangent line/derivative
Find the values of $\displaystyle x$ for all points on the graph of $\displaystyle f(x)= x^3 + 2x^2 + 5x -16$ at which the slope of the tangent line is 5.

my work:

$\displaystyle f(x)= x^3 + 2x^2 + 5x -16$
$\displaystyle f(x) = 3x^2 + 4x + 5$
$\displaystyle 5 = 3x^2 + 4x + 5$
$\displaystyle 0 = 3x^2 + 4x$

stuck after this...
• Oct 23rd 2008, 04:12 PM
skeeter
Quote:

Originally Posted by algebra2
Find the values of $\displaystyle x$ for all points on the graph of $\displaystyle f(x)= x^3 + 2x^2 + 5x -16$ at which the slope of the tangent line is 4.

my work:

$\displaystyle f(x)= x^3 + 2x^2 + 5x -16$
$\displaystyle f(x) = 3x^2 + 4x + 5$
$\displaystyle 5 = 3x^2 + 4x + 5$
$\displaystyle 0 = 3x^2 + 4x$

stuck after this...

$\displaystyle 5 = 3x^2 + 4x + 5$ ??? look again at your problem statement.

should be ...

$\displaystyle 4 = 3x^2 + 4x + 5$

$\displaystyle 0 = 3x^2 + 4x + 1$

$\displaystyle 0 = (3x + 1)(x + 1)$

finish.

• Oct 23rd 2008, 05:14 PM
algebra2
oh sorry i typed it wrong, the slope should be 5.

...and how do you factor polynomials with a number in front like $\displaystyle 3x^2 + 4x$, there's a three in front of the x^2. is there a technique to this?
• Oct 23rd 2008, 05:24 PM
skeeter
first rule of factoring ... pull out whatever factors each term has in common.

$\displaystyle 3x^2 + 4x = 0$

$\displaystyle x(3x + 4) = 0$

$\displaystyle x = 0$

$\displaystyle x = -\frac{4}{3}$
• Oct 23rd 2008, 05:28 PM
algebra2
i see, thank you.