Show that no line with a y-int of 10 will ever be tangential to the curve with $y=3x^2+7x-2$ Show that no line with a y-int of 10 will ever be tangential to the curve with $y=3x^2+7x-2$.
Having trouble in showing this. So far these are my process. 


*

*Let line be $y=mx+10$

*$mx+10 = 3x^2+7x-2$

*$3x^2+(7-m)x-12=0$

*Apply quadratic formula

*$\frac{(7-m\pm\sqrt{(m^2-14m+193)}}6$


A bit stuck here. Maybe I've missed the whole point and complicated this. Any help is appreciated! Thanks in advance :)

The slope of the tangent line at a point $x_0$ of $y=3x^{2}+7x-2$ is $6x_{0}+7$. Suppose, on contrary, that at the point $(x_{0},3x_{0}^{2}+7x_{0}-2)$, the tangent line has $y$-intercept $10$. Then, as you pointed out, the equation of this tangent line is $y=(6x_{0}+7)x+10$. Since this line is tangent to the curve $y=3x^2+7x-2$, it follows that 
$$
3x_{0}^{2}+7x_{0}-2=(6x_{0}+7)x_{0}+10\implies3x_{0}^{2}=-12\implies x_{0}^{2}=-4,
$$
which is impossible. Therefore, we have a contradiction and the result is proved.
