Find the equations of the tangent line(s) that are parallel to the line $y= -4x +3$ Find the equations of the tangent line(s) to $\;f(x) = x^3 -4x^2 + 2\;$ that are parallel to the line $y= -4x +3$.
I know the slope is -4 and $f'(x)= 3x^2-8x.$
 A: if $f'(x)=-4$ then the tangent line to that point will be parallel to your line.
Set $f'(x)=-4$ then solved for $x$.
$$-4=3x^2-8x$$
$$3x^2-8x+4=0$$
$$x=\dfrac{8\pm\sqrt{(-8)^2-4(3)(4)}}{2(3)}$$
$$x=\dfrac{8\pm\sqrt{64-48}}{6}$$
$$x=\dfrac{8\pm4}{6}$$
$$x=\dfrac{8+4}{6}\text{ or }\dfrac{8-4}{6}$$
$$x=2\text{ or }\dfrac23$$
Now for both of these points we find the equation of the line that is tangent to them.
For $x=2$:
$$f(x)=2^3-4(2)^2+2$$
$$=8-16+2$$
$$=-6$$
Now using the point slope form we can find the equation of the line tangent to that point:
$$y-y_1=m(x-x_1)$$
$$y-(-6)=-4(x-2)$$
$$y+6=-4x+8$$
$$y=-4x+2$$
That is the equation of the line that is tangent to $x=2$
For $x=\dfrac23$:
$$f(x)=\left(\dfrac23\right)^3-4\left(\dfrac23\right)^2+2$$
$$=\dfrac8{27}-\dfrac{16}{9}+2$$
$$=\dfrac{-40}{27}+2$$
$$=\dfrac{14}{27}$$
Now using the point slope form we can find the equation of the line tangent to that point:
$$y-y_1=m(x-x_1)$$
$$y-\left(\dfrac{14}{27}\right)=-4\left(x-\left(\dfrac23\right)\right)$$
$$y-\left(\dfrac{14}{27}\right)=-4x+\dfrac83$$
$$y=-4x+\dfrac{86}{27}$$
That is the equation of the line that is tangent to $x=\dfrac23$
So the equations of the two lines are:$$y=-4x+\dfrac{86}{27}$$ and $$y=-4x+2$$
