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the equation of the line tangent to the graph of y= tan(x)+3 at the point (0,3) is:
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If you take the derivative of y = tan (x) + 3 you get: y' = sec^2 (x) and the value at x = 0 is y' = 1, so that is your slope. Using the slope-intercept form for the equation of a line: y = mx + b we get: y = (1)x + 3 or y = x + 3
Here's a graph of the original equation and of the line:
All good now?
yep thanks
Good luck in all of your studies and thx for the recognition! @zombiejr
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