Find the equation of the tangent line to the curve y=1+ xe^(2x)
y'=(e^(2x))(2x+1) = slope
What's the next step?
at the point where x=0 *
substitute 0 in for the slope equation to get slope at x=0, then substitute 0 in for the original equation to get a point (0,y). Then use point slope equation y-y1=m(x-x1). m is the slope and x1,y1 is the point from substituting 0 into the original equation.
The next step is to find the x-coordinate where you want to determine the equation of the tangent line. This x-coordinate will help us in finding the corresponding y-coordinate as well as the slope of the tangent line.
Let's say the x-coordinate we are interested in is denoted as 'a'. We can substitute this value into the equation y = 1 + xe^(2x) to find the y-coordinate.
So, substitute x = a into the equation: y = 1 + ae^(2a).
Now, we have the coordinates (a, y) and the slope (y') that we found earlier. We can use the point-slope form of a line to write the equation of the tangent line.
The point-slope form is given by: y - y₁ = m(x - x₁), where (x₁, y₁) is a point on the line, m is the slope of the line.
So, using the coordinates (a, y) and the slope y', we can substitute them into the point-slope form:
y - y₁ = y'(x - x₁)
Replacing y₁ with the y-coordinate we found earlier and substituting x₁ with our chosen x-coordinate 'a', we can rewrite the equation as:
y - (1 + ae^(2a)) = (e^(2a))(2a + 1)(x - a)
Simplifying further will give us the equation of the tangent line.