What is the equation of the line tangent to f(x)=(x-4)^2-x  at x=-1?

May 18, 2017

$y = f \left(x\right) = - 11 x + 15$

Explanation:

$f \left(x\right) = {\left(x - 4\right)}^{2} - x = {x}^{2} - 9 x + 16$
at $x = - 1 , y = f \left(- 1\right) = {\left(- 5\right)}^{2} - \left(- 1\right) = 26$

gradient function, $f ' \left(x\right) = 2 x - 9$, therefore it gradient at $x = - 1$,
$f ' \left(- 1\right) = 2 \left(- 1\right) - 9 = - 11$

with using standard form of line equation, $f \left(x\right) = y = m x + c$, plug in $x = - 1 , y = 26 \mathmr{and} m = - 11$, into this equation to find $c$

$26 = - 11 \left(- 1\right) + c$
$15 = c$

therefore it line tangent,
$y = f \left(x\right) = - 11 x + 15$