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 :)
 A: You tagged calculus so with derivatives: the slope of a tangent to the given function is
$$y'=6x+7\implies\;\text{for any point on the graph }\;\;(a, 3a^2+7a-2)$$
the tangent line to the function at that point is
$$y-(3a^2+7a-2)=(6a+7)(x-a)\implies y=(6a+7)x-3a^2-2$$
and thus the $\;y\,-$ intercept is $\;-3a^2-2\;$ , and this is $\;10\;$ iff
$$-3a^2-2=10\iff a^2=-4$$
and this last equality can't be true in the real numbers
A: You got (almost)
$$x_{1,2}={m-7\pm\sqrt{(m-7)^2+144}\over6}\ .$$
As $(m-7)^2+144>0$ for all $m$ this says that  you obtain two different points of intersection whatever the chosen slope $m$ is.
If the discriminant would have been, e.g., $m^2-5m+6$ then there would be two "special" values of $m$ for which the discriminant is $0$. The two lines with these slopes would then be tangents to the parabola.
A: 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.
A: There are no tangents to a parabola from a point in its “interior.” This parabola opens upwards, i.e., in the positive $y$ direction and intersects the $y$-axis at $y=-2$, so the point $(0,10)$ is in the interior and therefore no tangent can pass through it.
A: If you assume that the line is tangent to the parabola then
$$
mx+10=3x^2+7x-2
$$
at some point. Simplifying the previous equation we have
$$
3x^2+(7-m)x-12=0.
$$
Since we are assuming they are tangential, the discriminant of the last equation should be equal to zero. That is,
$$
m^2-14m+193=0.
$$
Solving for $m$ using the quadratic formula you get 
$x=7+12i$ or $x=7-12i$, which are not solutions in the real numbers. Therefore the line and the parabola are not tangential.
