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Write the equation of the tangent to the curve at point x
I'm studying a textbook for self-study and it provides an answer to this question on page 86 of the book. The question is as follows:
Write the equation of the tangent to the curve at point $x=0$
Here is the answer provided in the book:
$\frac{y-y_0}{x}=-1$
This is an equation of a line through the origin.
What confuses me is the statement:
Thus, the equation of the tangent at the origin is $y=y_0-x$
Why is it that the $-x$? I thought it was by definition that the tangent to the curve at the point on the curve $x$ that is tangent to the curve $y$ at the point is the curve $y=-x$.
I think that my confusion comes from the fact that it is tangent to the curve at the origin. If that's the case, would the $-x$ be the tangent to the curve at the origin? Or is the $-x$ in the original equation, $y-y_0$, meant to denote that the $-x$ is tangent to the curve at the origin?
A:
It's all about how you interpret the equation of a line through the origin.
The usual interpretation is as the equation of a line tangent to the given curve. The equation
$y-y_0=-x$
is the same as the equation
$y+y_0=x.$
The function $y+y_0$ is increasing, while $x$ is decreasing. This shows that the curve is concave upward, and hence the tangent line is concave downward.
There is an alternative interpretation as the equation of a line parallel to the $x$ axis. That is,
$y-y_0=-x$
is the same as the equation
$y=y_0-x.$
The function $y=y_0-x$ is increasing, while $ 0b46394aab
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