Question

Find the point(s) where the line tangent to the graph of:
f(x)=(16/9)x^3-2x

is perpendicular to the line: 2y-3x+(16/9)=0

Answer 1

did you know how to get derivatives??

Answer 2

@Matthew071

Answer 3

Yes

Answer 4

calculate for the first derivative of the function: f(x)=(16/9)x^3-2x

next, change the line: 2y-3x+(16/9)=0 into slope-int form to get the slope of this line

Answer 5

tell me what you got..

Answer 6

ok

Answer 7

So the derivitive of the function = \[(16/3)x^3 -2\]

Im not to sure how to get to slope intercept form

Answer 8

change the line into:\[y=mx+b\] where m-slope, b is the y-int
in other words, arrange the equation in terms of y