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

My answer: 3

1x+3x+2=6x

f(x) -f(1)/x(-1)=3

Is that right?? :confused:

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- Mar 26th 2006, 03:39 PMbatman123Checking Tangent
Find the slope and the equation of the tangent line to the function f(x)=x^2+3x+2 at x=1.

My answer: 3

1x+3x+2=6x

f(x) -f(1)/x(-1)=3

Is that right?? :confused: - Mar 26th 2006, 03:45 PMThePerfectHackerQuote:

Originally Posted by**batman123**

thus,

$\displaystyle f'(x)=2x+3$

thus,

$\displaystyle f'(1)=2(1)+3=5$

Thus, slope is 5.

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To find equation use the formula,

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

where $\displaystyle (x_0,y_0)$ is a point.

Notice if $\displaystyle x_0=1$ then, $\displaystyle y_0=f(1)=6$

Thus,

$\displaystyle y-6=5(x-1)$

Solve for $\displaystyle y$ thus,

$\displaystyle y=5x+1$ - Mar 26th 2006, 04:16 PMbatman123Thanks
thank you I see what I did wrong.