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Find the EQUATION OF THE TANGENT to g(x)=(x-4)^2 at it’s y-intercept. Please use an appropriate limit, much appreciated!

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Answer: y = –8x + 16

Step-by-step explanation

Given


g\mleft(x\mright)=\left(x-4\right)^2

First, we have to know the y-intercept, which is the point where x = 0:


g(0)=(0-4)^2
g(0)=16

Thus, the y-intercept is (0, 16).

Next, we have to get the derivative of the function, as the tangent line is the derivative:


(dg(x))/(dx)=2(x-4)

Then, we have to compute the slope:


m=-8

Finally, we find the line in the slope-intercept form (y = mx + b):


b=16

Thus:


y=-8x+16

User Ineersa
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