tangent line/derivative

**algebra2** · October 23rd 2008, 04:04 PM

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

my work:

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

stuck after this...

**skeeter** · October 23rd 2008, 04:12 PM

Originally Posted by algebra2

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

my work:

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

stuck after this...

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

should be ...

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

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

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

finish.

**algebra2** · October 23rd 2008, 05:14 PM

oh sorry i typed it wrong, the slope should be 5.

...and how do you factor polynomials with a number in front like $3x^2 + 4x$ , there's a three in front of the x^2. is there a technique to this?