i know that you can use the point slope formula. but what if all they give for is x=1, for y=e^(-x )+ e^(2x-1). do you find the derv of the function and then plug in 1 to X to find y and then use the point slope formula?
i know that you can use the point slope formula. but what if all they give for is x=1, for y=e^(-x )+ e^(2x-1). do you find the derv of the function and then plug in 1 to X to find y and then use the point slope formula?
If y=e^(-x)+e^(2x-1) then no that's not the derivative. The second term is wrong. It should be (2)e^(2x-1). Do you see why?
Once you get the right y', plug in x=1 to get the slope of y at x=1. Then use the point-slope formula: $\displaystyle y-y_0=m(x-x_0)$ to find the tangent line.