## Calculus: Early Transcendentals 8th Edition

(a) $y = \frac{1}{4}x +\frac{ln(4)+1}{4}$ (b) $y = e~x$
(a) $x-4y = 1$ $4y = x-1$ $y = \frac{1}{4}x-\frac{1}{4}$ If the tangent line is parallel to this line, then the slope of the tangent line must have the same slope of $\frac{1}{4}$ We can find the point on the curve $y = e^x$ where the slope is $\frac{1}{4}$: $y = e^x$ $y' = e^x = \frac{1}{4}$ $ln~(e^x) = ln(\frac{1}{4})$ $x = ln(\frac{1}{4})$ $x = -ln(4)$ When $x = ln(\frac{1}{4}),~~$ then $~~y = e^{ln(1/4)} = \frac{1}{4}$ The tangent line meets the curve $y = e^x$ at the point $(-ln(4), \frac{1}{4})$ We can find the equation of the tangent line: $y - \frac{1}{4} = \frac{1}{4}[x - (-ln(4)]$ $y = \frac{1}{4}x +\frac{1}{4}~ln(4)+\frac{1}{4}$ $y = \frac{1}{4}x +\frac{ln(4)+1}{4}$ (b) The tangent line meets the curve at the point $(x,e^x)$ If the tangent line passes through the origin, then the slope of the line is $m = \frac{e^x-0}{x-0} = \frac{e^x}{x}$ Also: $m = y' = e^x$ Then $~~e^x = \frac{e^x}{x},~~$ and so $~~x = 1$ The slope of the tangent line is $m = e^1 = e$ When $x = 1~~$ then $~~y = e$ The tangent meets the curve at the point $(1,e)$ We can find the equation of the tangent line: $(y-e) = e(x-1)$ $y = e~x-e+e$ $y = e~x$