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How do you write the slope of the line tangent to #g(x)=x^2-4# at the point (1,-3)?

Answer 1

Cameron Alexander

First, we know that #dy/dx# is the slope of the curve at any point. So, we can find #g'(x)# at the beginning.

#g'(x)=d/dx(x^2-4)=d/dx(x^2)-d/dx(4)=2x-0=2x#

Then, at the point #(1,-3)# , the slope of the line tangent to #g(x)# is just #g'(1)#, by replacing the value of #x#.

Therefore, the slope of tangent at point #(1,-3)# #=g'(1)=2(1)=2#

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Answer 2

Adam Adkins

To find the slope of the line tangent to the function g(x) = x^2 - 4 at the point (1, -3), we can use the derivative of the function. The derivative of g(x) is given by g'(x) = 2x.

To find the slope at the point (1, -3), we substitute x = 1 into the derivative equation: g'(1) = 2(1) = 2.

Therefore, the slope of the line tangent to g(x) = x^2 - 4 at the point (1, -3) is 2.

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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.