Question

Help with finding the equation of a line tangent to a graph.

Answer 1

\[y = \ln x + \sin x\]

Answer 2

How do I start?

Answer 3

@satellite73

Answer 4

@whpalmer4

Answer 5

Take the derivative of \(y\) with respect to \(x\). Evaluate it at the point where you want the tangent line to touch the graph. That gives you the slope of the tangent line. Then find the correct \(y\)-intercept so that the tangent line actually touches the graph at that point.

Example: find the equation of a tangent line to \(y=3x^2\) at \(x = 1\)

\[y = 3x^2\]\[\frac{dy}{dx} = 6x\]evaluate that at \(x=1\):
\[m = \frac{dy}{dx} = 6(1) = 6\]Slope of the tangent line is \(m = 6\)
Now we need to find the value of \(y = f(1)\) to get the \(y\)-intercept:
\[y = f(1) = 3(1)^2 = 3\]Tangent line has slope \(m = 6\) and passes through point \((1,3)\) so equation of tangent line is
\[y = 6x+ b\]\[3=6(1)+b\]\[b=-3\]\[y =6x-3\]

Plot showing function and tangent at \(y = 1\) attached.