Tangent to curve y = e^x + 1 at P parallel to y = 3x + 1

[Monkeyseat]

Question: A curve has equation y = e^x +1. The tangent to the curve at the point P is parallel to the line y = 3x + 1. Find, in an exact form, the coordinates of P.

Working: I'm not sure where to go with this. I know I need to find the tangent of y = e^x +1.

y = e^x +1
dy/dx = e^x

Sorry for the short amount of working, but I don't know where to go from there. :? I don't know how to find the coordinates/gradient. Any help greatly appreciated.

Thanks.

 

[stapel]

I don't know how to find the...gradient.

You might want to review the definition of the derivative, especially the value of the derivative at a point, and think about the relationship between the slope of a straight line (such as a tangent line) and the derivative at a point. Then think back to algebra, where you learned about slopes of parallel and perpendicular lines.... :wink:

Eliz.

 

[galactus]

Since the tangent to the curve is parallel to y=3x+1, it must have the same slope which is m=3.

The slope is \(\displaystyle e^{x}\). What point solves \(\displaystyle e^{x}=3\)?.

 

[Monkeyseat]

D'oh! I didn't realise it was so simple. I've done it now.

x = ln 3, y = 4.

Cheers.