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

Answer 1

Emilia Aguilar

#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]}

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Answer 2

Emma Aldridge

The equation of the line tangent to f(x)=x^2 +6x-9 at x=0 is y = -9.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.