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??
Slope is the derivative evaluated at $\displaystyle x=1$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,
$\displaystyle y=5x+1$