Find the equation of the tangent to y=2-xe^x at the point where x=0
So I found the first derivative by doing the sum rule and product rule.
I got e^(x)(x-1)
Then I found the slope by substituting 0 for x and got -1.
Then I found y by substituting 0 into the original equation and got 2.
So for my final equation after finding b I had y=-1x+2
But in the back of the book the answer says x+y-2=0.
You can simplify things right from the start by writing the equation for the tangent as:
Delta Y/DeltaX = y'
The derivative y' is to be evaluated at the point where you take the tangent, Delta Y and Delta x are the differences in the y and x coordinate relative to the values they take at the point where the tangent touches the graph of the function.
To find the equation of the tangent line to the curve y = 2 - xe^x at the point where x = 0, you correctly calculated the first derivative, dy/dx, as e^x(x - 1).
Next, evaluate the slope (m) of the tangent line by substituting x = 0 into the derivative:
dy/dx = e^0(0 - 1) = -1
So, the slope m of the tangent line is -1.
To find the y-coordinate (y-value) of the point where x = 0 on the curve, substitute x = 0 into the original equation:
y = 2 - 0e^0 = 2
Hence, the point where x = 0 on the curve is (0, 2).
Now that we have the slope (m = -1) and a known point (0, 2), we can use the point-slope form of a linear equation to find the equation of the tangent line:
y - y1 = m(x - x1)
Plugging in the values: x1 = 0, y1 = 2, m = -1, we get:
y - 2 = -1(x - 0)
Simplifying:
y - 2 = -x
To write the equation in the standard form (ax + by + c = 0), bring all terms to one side:
x + y - 2 = 0
Therefore, the correct equation of the tangent line to the curve y = 2 - xe^x at the point where x = 0 is x + y - 2 = 0.