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

• Jan 26th 2010, 01:28 PM
bigwave
eq of tangent lines that pass thru pt to graph of f
find the equations of the tangent lines to \$\displaystyle f(x)=4x-x^2\$ as they both pass thru point \$\displaystyle (2,5) \$which is not on the graph.

\$\displaystyle
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 \$\displaystyle 4-2x \$
anyway not seeing how this these 2 equations are derived

appreceiate much..
• Jan 26th 2010, 01:53 PM
drumist
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 \$\displaystyle (x_0,y_0)\$:

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

\$\displaystyle \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 \$\displaystyle (2,5)\$:

\$\displaystyle 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 \$\displaystyle x_0\$ satisfy the conditions. Then we can get the two tangent lines at those two points.
• Jan 26th 2010, 02:16 PM
Quote:

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

\$\displaystyle
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 \$\displaystyle 4-2x \$
anyway not seeing how this these 2 equations are derived

appreceiate much..

Hi bigwave,

The function derivative gives the tangents' slope.

\$\displaystyle 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...

\$\displaystyle 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...

\$\displaystyle 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.
• Jan 26th 2010, 02:41 PM
bigwave
the 2 tangent line equations are:
thus

the 2 tangent line equations are:

\$\displaystyle
y=2x+1\$
\$\displaystyle
y=-2x+9\$