For which values of $m$ the line is tangent to the quadratic curve?

Question

For which values of $m$, the line $y=2x-4$ is tangent to the curve $y=(m+3)x^2+mx ?$

We have a quadratic equation. the equation of slope of tangent line to it for specific $x$ can be find by $y'=(2m+6)x+m$. but when I compare it to $y=2x-4$, I can't find $m$.

Answer 1

You are after a point $(a,b)\in\Bbb R^2$ such that:

• $(a,b)$ belongs to the curve $y=(m+3)x^2+mx$;
• $(a,b)$ belongs to the line $y=2x-4$;
• the slope of $y$ at $(a,b)$ is equal to $2$.

This means that$$\left\{\begin{array}{l}b=(m+3)a^2+ma\\b=2a-4\\(2m+6)a+m=2\end{array}\right.$$There are only two solutions: $m=-2$ (in which case $a=2$ and $b=0$) and $m=22$ (in which case $a=-\frac25$ and $b=-\frac{24}5$).

Answer 2

$m$ can be found without using slope. The given line touches the curve at $(x,2x-4)$ so that $$2x-4=(m+3)x^2+mx$$ is satisfied. This is a quadratic in $x$, $$(m+3)x^2+(m-2)x+4=0$$ and its discriminant is zero, since for a given $m$, the quadratic has a unique root (tangency point). Thus $$D : (m-2)^2-4\cdot 4 \cdot (m+3)=0$$ $$\Rightarrow m^2-20m-44=0 \Rightarrow m=22,-2$$