At what point on the given curve is the tangent line parallel to the line 3x - y = 2? y = 3 + 2ex − 5x How do i do this?
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Originally Posted by rhcp1231 At what point on the given curve is the tangent line parallel to the line 3x - y = 2? y = 3 + 2ex − 5x How do i do this? You need to set the derivative of the curve to the slope of the given line and solve for x. By the way is 2e^x or 2ex?
2e^x so the derivative is 2e^x-5, how do i solve for x for that?
Originally Posted by rhcp1231 2e^x so the derivative is 2e^x-5, how do i solve for x for that? $\displaystyle 3=2e^x-5$ $\displaystyle 8=2e^x$ $\displaystyle e^x=4$ $\displaystyle x=ln(4)$ Then just plug that into the original equation for the curve and you have the point where te tangent line is parallel to the given line.
Originally Posted by Keithfert488 $\displaystyle 3=2e^x-5$ $\displaystyle 8=2e^x$ $\displaystyle e^x=4$ $\displaystyle x=ln(4)$ and the 3 comes from 3x - y = 2 right?
Originally Posted by rhcp1231 and the 3 comes from 3x - y = 2 right? Yes. If you put the equation into slope intercept format (y=mx+b) you find that y=3x-2 so the slope is three.
Originally Posted by Keithfert488 Yes. If you put the equation into slope intercept format (y=mx+b) you find that y=3x-2 so the slope is three. sweet thanks for the help
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