Answer
$y=32x-47$
Work Step by Step
Let $g:32x-y=15$ and $l$ be the tangent line to the curve $y=x^4+1$.
Find the slope of $g$, $m_g$:
$32x-y=15$
$y=32x-15$
$m_g=32$
Find $\frac{dy}{dx}$:
$\frac{dy}{dx}=\frac{d}{dx}(x^4+1)=4x^3$
Find the values of $x$ such that $g$ is parallel to $l$ or $\frac{dy}{dx}=m_g$:
$4x^3=32$
$x^3=8$
$x=2$
Find the equation of $l$:
$y-y(2)=m_g(x-2)$
$y-(2^4+1)=32(x-2)$
$y-17=32x-64$
$y=32x-47$
Thus, the tangent line to the curve $y=x^4+1$ and parallel to $32x-y=15$ has the equation $y=32x-47$.
