The equation of the tangent line is $y-f(a)=f^{\prime}(a)(x-a)$.Bushy said:The function $f(x)= e^x+k$ has a tangent to the curve at $x=a$ and going through the origin. Find $k$ in terms of $a$
A tangent line is a line that touches a curve at only one point. It represents the instantaneous rate of change of the curve at that point.
The slope of a tangent line is equal to the derivative of the curve at the point of tangency. In this case, the derivative of $e^x+k$ is simply $e^x$, so the slope of the tangent line is $e^a$.
The equation of a tangent line can be found using the point-slope form: $y-y_1=m(x-x_1)$, where $m$ is the slope and $(x_1,y_1)$ is the point of tangency. In this case, the equation would be $y-(e^a+k)=e^a(x-a)$.
The y-intercept of a tangent line can be found by substituting the x-coordinate of the point of tangency into the equation of the tangent line. In this case, the y-intercept would be $e^a(a+1)+k$.
Finding the tangent to a curve allows us to approximate the behavior of the curve at a specific point. It can also be used to find the maximum and minimum points of a curve, as well as its concavity and points of inflection.