Find the equations of all tangent lines to the function $f(x)=x^{2}+4x-2$ that go through the point $(5,2)$

Question

I have tried a few things. The tangent line does not go through the origin so there must be a value of b (in $y=mx+b$). I am not sure where to go from where I started which is finding the general equation of the derivative of $f(x)$.

Here is what I have tried:

I know that m (the slope) is equal to $2x + 4$.

From then I set two equations to equal each other (that I found by using the point slope formula of a line that goes through $(5,2)$.

Answer 1

$\newcommand{\R}{\mathbb{R} }$ Given $f:I \subseteq \R \rightarrow \R$ a differentiable function, we know that an equation of a tangent line in $p \in I $ is $ y=f(p)+f'(p)(x-p) \; . $

Then, you have to find $p \in I$ such that the equation $ 2=f(p)+f'(p)(5-p) $ is satisfied.

Have fun!

Answer 2

Another method:

Take a line through $(5,2)$: $\;y=2+t(x-5)$, and the equation for the intersection points with the parabola: $\;y=x^2+4x-2=2+t(x-2)$, whence the equation with parameter: $$x^2+(4-t)x+2(t-2)=0.$$ The line with slope $t$ is tangent to the curve if this equation has a double root, which is equal to $\dfrac{t-4}2$, i.e. if $$\Delta=(4-t)^2-8(t-2)=0.$$ As this is a quadratic equation for $t$, you'll find two lines tangent to the parabola.