# At what points on the graph of f(x)=2x^3-9x^2-12x+5 is the slope of the tangent line 12?

Apr 13, 2015

The derivative of a function gives us the slope of a tangent line for a specified value of $x$

$f \left(x\right) = 2 {x}^{3} - 9 {x}^{2} - 12 x + 5$
implies
$f ' \left(x\right) = 6 {x}^{2} - 18 x - 12$

and we are asked to find when this is $= 12$

$6 {x}^{2} - 18 x - 12 = 12$

${x}^{2} - 3 x - 4 = 0$

which factors as
$\left(x - 4\right) \left(x + 1\right) = 0$

when $x = 4$
$f \left(x\right) = 2 {\left(4\right)}^{3} - 9 {\left(4\right)}^{2} - 12 \left(4\right) + 5$
$= - 59$

when $x = - 1$
$f \left(- 1\right) = 2 {\left(- 1\right)}^{3} - 9 {\left(- 1\right)}^{2} - 12 \left(- 1\right) + 5$
$= 6$

So $f \left(x\right)$ has a tangent with slope $12$ at
$\left(4 , - 59\right)$
and
$\left(- 1 , 6\right)$