# Thread: eq of tangent lines that pass thru pt to graph of f

1. ## eq of tangent lines that pass thru pt to graph of f

find the equations of the tangent lines to $f(x)=4x-x^2$ as they both pass thru point $(2,5)$which is not on the graph.

$
f'(x) = 4-2x
$

but we do not the points on the graph where the 2 lines from this point will be tangent. we know the slope of line will be same as $4-2x$
anyway not seeing how this these 2 equations are derived

appreceiate much..

2. There might be an easier way, but this is how I would tackle this problem:

First we can find a general form for a tangent line at the point $(x_0,y_0)$:

$y-y_0=m(x-x_0)$

$\implies y-(4x_0-{x_0}^2)=(4-2x_0)(x-x_0)$

Now we narrow it down to ones that pass through the point $(2,5)$:

$5-(4x_0-{x_0}^2)=(4-2x_0)(2-x_0)$

This should simplify to a quadratic equation that we can solve to find what values of $x_0$ satisfy the conditions. Then we can get the two tangent lines at those two points.

3. Originally Posted by bigwave
find the equations of the tangent lines to $f(x)=4x-x^2$ as they both pass thru point $(2,5)$which is not on the graph.

$
f'(x) = 4-2x
$

but we do not the points on the graph where the 2 lines from this point will be tangent. we know the slope of line will be same as $4-2x$
anyway not seeing how this these 2 equations are derived

appreceiate much..
Hi bigwave,

The function derivative gives the tangents' slope.

$m=4-2x$

At the points of tangency, the y co-ordinate (and x co-ordinate) is the same on both the tangents and the curve.

(2,5) is on the tangents but not the curve...

$y-5=m(x-2)=(4-2x)(x-2)\ \Rightarrow\ y=5+4x-8-2x^2+4x=-2x^2+8x-3$

y also satisfies the curve equation...

$4x-x^2=-2x^2+8x-3\ \Rightarrow\ x^2-4x+3=0$

The x co-ordinates of the points of tangency can be found.
These can be used to write the exact slopes and equations of the tangents.

4. ## the 2 tangent line equations are:

thus

the 2 tangent line equations are:

$
y=2x+1$

$
y=-2x+9$