Answer
$y=-9x$
$y=-\frac{9}{4}x-\frac{27}{4}$
Work Step by Step
$y=x^3-9x$
$\frac{dy}{dx}=3x^2-9$
$(y-y_1)=m(x-x_1)$
$(x_1, y_1)=(1,-9)$
$(y-(-9))=m(x-1)$
$y+9=m(x-1)$
$y=m(x-1)-9$
$y=(3x^2-9)(x-1)-9$
$(x^3-9x)=(3x^2-9)(x-1)-9$
$x^3-9x=3x^3-3x^2-9x+9-9$
$2x^3=3x^2$
$2x^3-3x^2=0$
$x^2(2x-3)=0$
$x=0$ or $\frac{3}{2}$
$x=0$
$y=(0)^3-9(0)$
$y=0$
$\frac{dy}{dx}=3x^2-9$
$3(0)^2-9=-9$
$m=-9$
$y=m(x-1)-9$
$y=(-9)(x-1)-9$
$y=-9x$
$x=\frac{3}{2}$
$y=(\frac{3}{2})^3-9(\frac{3}{2})$
$y=\frac{27}{8}-\frac{27}{2}$
$y=-\frac{81}{8}$
$\frac{dy}{dx}=3x^2-9$
$3(\frac{3}{2})^2-9=\frac{27}{4}-9=-\frac{9}{4}$
$m=-\frac{9}{4}$
$y=m(x-1)-9$
$y=(-\frac{9}{4})(x-1)-9$
$4y=-9(x-1)-36$
$4y=-9x+9-36$
$4y=-9x-27$
$y=-\frac{9}{4}x-\frac{27}{4}$
