What is the equation of the line tangent to #f(x)=x^2 +6x-9 # at #x=0#?

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Answer 1 · 2016-01-23T20:03:55

#y=6x-9#

Find the point the tangent line will intercept.

#f(0)=-9#

The tangent line will pass through the point #(0,-9)#.

To find the slope of the tangent line, find the value of at the derivative at #x=0#.

To find the derivative, use the power rule.

#f(x)=x^2+6x-9#

#f'(x)=2x+6#

The slope of the tangent line at #x=0# is #f'(0)#.

#f'(0)=6#

We know the tangent line passes through the point #(0,-9)# and has slope #6#.

We can relate these in an equation in point-slope form, which is

#y-y_1=m(x-x_1)#

Where you know a point #(x_1,y_1)# on the line and the slope #m# of the line.

Thus, the line's equation is

#y-(-9)=6(x-0)#

Which, rewritten, is

#y=6x-9#

Graphed are the function and the tangent line:

graph{(x^2+6x-9-y)(y-6x+9)=0 [-20, 20, -30.45, 28.05]}