Answer:
[tex]\huge\boxed{(0,4)}[/tex]
[tex]\huge\boxed{\left(\frac{2}{3},\ \frac{32}{27}\right)}[/tex]
Step-by-step explanation:
We are solving for the points on the curve where the tangent line has a slope of 4.
In other words, we are solving for the points when [tex]y' = 4[/tex].
First, we can expand the function.
[tex]y=(x-1)(x^2-4)[/tex]
[tex]y= x^3 - 4x - x^2 + 4[/tex]
[tex]y=x^3-x^2-4x+4[/tex]
Next, we can take its derivative using the power rule.
↓↓↓
[tex]y' = 3x^2 - 2x - 4[/tex]
Now, we can set y' to 4 and solve for x to get the x-coordinates of the points when the curve is parallel to y = 4x:
[tex]4 = 3x^2-2x-4[/tex]
[tex]0 = 3x^2 - 2x[/tex]
[tex]0=x(3x-2)[/tex]
⇒ [tex]x = 0[/tex] or [tex]3x - 2 = 0[/tex]
[tex]x = 2/3[/tex]
Finally, we can plug these x-coordinates back into the original function to get points:
1) [tex]y = (0 - 1)(0^2 - 4)[/tex]
[tex]y = (-1)(-4)[/tex]
[tex]y=4[/tex]
⇒ [tex]\huge\boxed{(0,4)}[/tex]
2) [tex]y=(2/3 - 1)((2/3)^2 - 4)[/tex]
[tex]y = (-1/3)(-32/9)[/tex]
[tex]y = 32/27[/tex]
⇒ [tex]\huge\boxed{\left(\frac{2}{3},\ \frac{32}{27}\right)}[/tex]
