Answer
Given, $y=6x^3+5x-3$. A tangent line with slope $4$ must be drawn at a point on the curve, which has the same slope, that is, $4$. For finding the slope of $y$, let's differentiate it with respect to $x$,
$$\frac{dy}{dx} = 18x^2+5$$This slope needs to be equal to $4$. Thus,
$$18x^2+5=4\implies18x^2+1=0$$$$x^2=-\frac{1}{18}$$No real value satisfies the above equation, hence the given curve has no tangent line with slope $=4$.
Work Step by Step
Given, $y=6x^3+5x-3$. A tangent line with slope $4$ must be drawn at a point on the curve, which has the same slope, that is, $4$. For finding the slope of $y$, let's differentiate it with respect to $x$,
$$\frac{dy}{dx} = 18x^2+5$$This slope needs to be equal to $4$. Thus,
$$18x^2+5=4\implies18x^2+1=0$$$$x^2=-\frac{1}{18}$$No real value satisfies the above equation, hence the given curve has no tangent line with slope $=4$.