What is the equation of the line tangent to # f(x)=sqrt(x^2+3x+6) # at # x=-1 #?

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Answer 1 · 2016-05-28T19:42:37

#y-2=1/4(x+1)#

We can find the point of tangency:

#f(-1)=sqrt(1-3+6)=sqrt4=2#

The function passes through the point #(-1,2)#.

Now, all we need to do is find the slope of the tangent line, which equals #f'(-1)#.

To find #f'(x)#, write the function using exponential powers.

#f(x)=(x^2+3x+6)^(1/2)#

Now, differentiate using the chain rule.

#f'(x)=1/2(x^2+3x+6)^(-1/2)*d/dx(x^2+3x+6)#

#f'(x)=1/(2(x^2+3x+6)^(1/2))*(2x+3)#

#f'(x)=(2x+3)/(2sqrt(x^2+3x+6))#

The slope of the tangent line is

#f'(-1)=(-2+3)/(2sqrt(1-3+6))=1/(2sqrt4)=1/4#

The equation of the line passing through #(-1,2)# with a slope of #1/4#, written in point-slope form, is

#y-2=1/4(x+1)#

Graphed are the function and its tangent line at #x=-1#:

graph{(y-sqrt(x^2+3x+6))(y-2-1/4(x+1))=0 [-10.625, 9.375, -1.72, 8.28]}