Find an equation of the tangent line to the curve at the point (-1,1) if y=4x^cubed +5x
tan = y/x tan = -1/1
\[y=4x^3+5x\] \[y'=12x^2+5\] 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
\[y-1=17(x+1)\]
how did you get from y=4x^3 +5x to y=12x^2+5?
he took the derivative of y = 4x^3+5x to get y' = 12x^2+5
what about synthetic division?
no you cannot use synthetic division to get the derivative
I already told you the answer is 3.141592654
they want an equation, not just the slope
besides, the answer isn't pi
?
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
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
answer should be \[\large y=17x+18\]
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