# Checking Tangent

• Mar 26th 2006, 04:39 PM
batman123
Checking Tangent
Find the slope and the equation of the tangent line to the function f(x)=x^2+3x+2 at x=1.

1x+3x+2=6x
f(x) -f(1)/x(-1)=3
Is that right?? :confused:
• Mar 26th 2006, 04:45 PM
ThePerfectHacker
Quote:

Originally Posted by batman123
Find the slope and the equation of the tangent line to the function f(x)=x^2+3x+2 at x=1.

1x+3x+2=6x
f(x) -f(1)/x(-1)=3
Is that right?? :confused:

Slope is the derivative evaluated at $x=1$
thus,
$f'(x)=2x+3$
thus,
$f'(1)=2(1)+3=5$
Thus, slope is 5.
-------
To find equation use the formula,
$y-y_0=m(x-x_0)$
where $(x_0,y_0)$ is a point.
Notice if $x_0=1$ then, $y_0=f(1)=6$
Thus,
$y-6=5(x-1)$
Solve for $y$ thus,
$y=5x+1$
• Mar 26th 2006, 05:16 PM
batman123
Thanks
thank you I see what I did wrong.