Show that no line with a y-int of 10 will ever be tangential to the curve with $y=3x^2+7x-2$

Question

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.

1. Let line be $y=mx+10$
2. $mx+10 = 3x^2+7x-2$
3. $3x^2+(7-m)x-12=0$
4. Apply quadratic formula
5. $\frac{(7-m\pm\sqrt{(m^2-14m+193)}}6$

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

Answer 1

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.

Answer 2

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

Answer 3

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.

Answer 4

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.