Answer
The equation of the tangent line that needs to be found is $$(l_0):y=32x-47$$
Work Step by Step
$$y=x^4+1$$
1) First, find the derivative $y$, or $y'$ $$y'=\frac{d}{dx}(x^4)+\frac{d}{dx}(1)$$
$$y'=4x^3$$
2) The slope of the tangent line $(l)$ to the curve $y$ at point $A(a,b)$ would be $y'(a)$.
Consider the mentioned line $(d)$: $$(d): 32x-y=15$$ $$(d):y=32x-15$$
The slope of line $(d)$ equals $32$. If we want the tangent line $(l)$ to be parallel with line $(d)$, both of their slopes must be equal. In other words, $$y'(a)=32$$ $$4a^3=32$$ $$a^3=8$$ $$a=2$$
So, $b=2^4+1=17$
Therefore, the tangent line $(l_0)$ at point $A(2,17)$ is parallel with line $(d)$.
The equation of tangent line $(l_0)$ is $$(l_0):(y-b)=y'(a)(x-a)$$ $$(l_0):y-17=32(x-2)$$ $$(l_0):y-17=32x-64$$ $$(l_0):y=32x-47$$
