Find an equation for the tangent line to the given curve at the point where x=x[0].
Give your answer in Y=mx+b form, where m and b are simplified fractions.
y=(4x)/(6x+10) ; x[0]=1

Please I really want the answer! thanks!

Question

Find an equation for the tangent line to the given curve at the point where x=x[0].
Give your answer in Y=mx+b form, where m and b are simplified fractions.
y=(4x)/(6x+10) ; x[0]=1

Please I really want the answer! thanks!

anonymous · Answer

$$y(x)=\frac{4x}{6x+10}\x_0=1$$$$y'(x)=\frac{4(6x+10)-6(4x)}{(6x+10)^2}\\ \ \ \ \ \ \ \ =\frac{24x+40-24x}{(6x+10)^2}\\ \ \ \ \ \ \ \ =\frac{40}{36x^2+120x+100}\\ \ \ \ \ \ \ \ =\frac{10}{9x^2+30x+25}\\ \ \ \ \ \ \ \ =\frac{10}{(3x+5)^2}\y'(x_0)=y'(1)=\frac{10}{[3(1)+5]^2}=\frac{10}{64}=\frac5{32}\y(x_0)=\frac{4(1)}{6(1)+10}=\frac4{16}=\frac14\Y-y(x_0)=y'(x_0)[x-x_0]\Y-\frac14=\frac5{32}[x-1]\Y=\frac5{32}x-\frac5{32}+\frac14=\frac5{32}x+\frac3{32}$$

anonymous · Answer

OMG!! Thank you so muchh!!! ;)