Find the equations of the tangent lines to curve $y=x^3+2x+1$ which are parallel to the line that passes through the points $(1,2)$ and $(3, 30)$.
What I have so far is,
Slope of the given line: $m=\frac{\left(30-2\right)}{\left(3-1\right)}=14$
Derivative of given curve $y'=3x^2+2$
Then I set the derivate equal to the slope that I found: $3x^2+2=14$, which gives us $x=±2$
I found the y-coordinates by plugging in the values of x to the equation of the curve: if $x=2$ then $y=13$ and if $x=-2$ then $y=-11$, and so the tangency points are: $(2,13)$ and $(-2,-11)$.
Then plugging in this information to the equation of a line $(y−y_0)=m(x−x_0)$ we get two tangents: $y=14x-15$ and $y=14x+17$.
Drawing the lines and the curve on GeoGebra shows that the lines are tangents to the curve, but they also cut through the curve. I thought that tangents aren't supposed to cut the curve at any point? Did I go wrong somewhere or can tangents cut the given curve as well? Sorry, I'm still pretty much in learning phase.