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Let be $f(x) = x^2 - 4x + 5$ and the point $A = (1, -2)$. I would like to find is there any tangent line to $f(x)$ which passes thru point $A$. If yes, what are the equation(s) ?

Thanks !

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    You have the slope using derivative, and a point on line (A). the rest would be easy. – 2017-01-23
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    Seriously, I can't find it how. I might miss something... – 2017-01-23

1 Answers 1

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Let's take a line passing through $A$ with slope $m$. Its equation is

$$y+2=m(x-1)$$

It intersects the parabola defined by $f(x)$ when $y=f(x)$. The abscissa of an intersection is therefore solution of the quadratic

$$x^2-(m+4)x+m+7=0$$

Now if this quadratic has no real solution the line doesn't intersect the parabola.

If it has two real solutions the line intersects at two distinct points

If it has one (double) root the line is tangent.

This is the case when

$$\Delta=(m+4)^2-4(m+7)=m^2+4m-12=0$$

this gives $m=2$ or $m=-6$. The equation of the two tangent to the parabola through $A$ are $y=2x-4$ and $y=-6x+4$.

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    I found c = m+7. – 2017-01-23
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    You're absolutely right. I edit – 2017-01-23