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Mathematics 8 Online
OpenStudy (anonymous):

Find an equation of the tangent line to the curve at the point (-1,1) if y=4x^cubed +5x

OpenStudy (anonymous):

tan = y/x tan = -1/1

OpenStudy (anonymous):

\[y=4x^3+5x\] \[y'=12x^2+5\] you want the slope at x = -1, get \[12(-1)^2+5=12+5=17\] so slope is 17, point is (-1,1) use point slope formula

OpenStudy (anonymous):

\[y-1=17(x+1)\]

OpenStudy (anonymous):

how did you get from y=4x^3 +5x to y=12x^2+5?

jimthompson5910 (jim_thompson5910):

he took the derivative of y = 4x^3+5x to get y' = 12x^2+5

OpenStudy (anonymous):

what about synthetic division?

jimthompson5910 (jim_thompson5910):

no you cannot use synthetic division to get the derivative

OpenStudy (anonymous):

I already told you the answer is 3.141592654

jimthompson5910 (jim_thompson5910):

they want an equation, not just the slope

jimthompson5910 (jim_thompson5910):

besides, the answer isn't pi

OpenStudy (anonymous):

?

OpenStudy (anonymous):

you have to use the derivative to find the slope. there is no other way to find the slope of a tangent line that i know of. it doesn't have a definition with out the derivative

OpenStudy (anonymous):

I figured it wasn't pie. im using the Slope = the limit of x of approaches a to use the formula f(x) -f(a)/ x-a..... the answer is y=7x+8

jimthompson5910 (jim_thompson5910):

answer should be \[\large y=17x+18\]

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