Find the tangent to the curve y=3x2+x+4 at x=3

The correct option is 19 (which represents the slope of the tangent to the curve).

To find the slope of the tangent to the curve $y=3x^2+x+4$ at $x=3$, we need to find the derivative of the function with respect to $x$ and then evaluate it at the given point.

First, let's find the first derivative, $\frac{dy}{dx}$:
$$\frac{dy}{dx}=\frac{d}{dx}(3x^2+x+4)$$
Using the power rule of differentiation, $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$, we get:
$$\frac{dy}{dx}=3·\left(2x\right)+1+0$$
$$\frac{dy}{dx}=6x+1$$

Next, we substitute the value $x=3$ into the derivative to find the slope of the tangent line at that point:
$${\left.\frac{dy}{dx}\right|}_{x=3}=6\left(3\right)+1$$
$${\left.\frac{dy}{dx}\right|}_{x=3}=18+1=19$$

Therefore, the slope of the tangent to the curve at $x=3$ is 19.