Question

The point on the curve y=2x^3 - 33x^2 + 180x +1 whre the tangent line to it is parallel to the line joining (2,13) and (1,1)?

Answer 1

find dy/dx ....
equate it to slope ... solve for x

Answer 2

find m, the slope of line through (2,13) and (1,1)

y=2x^3 - 33x^2 + 180x +1

\[\frac{dy}{dx} = 6x^2 - 66x + 180 = m\]

which is a quadratic.

is this possible without calculus? @experimentX

Answer 3

not sure ... if you like to try ... i'm ready to assist!!

Answer 4

ahh unfortunately i must return to physics revision :( exam monday

Answer 5

seems possible ...
f(x1) - f(x2) = (x1 - x2)(some polynomial of x)

Answer 6

i dint understand

Answer 7

first i have to find the slope by y2-y1/x2-x1

Answer 8

then dy/dx=0

Answer 9

then wat to do?

Answer 10

taking limit, we can replace
(f(x1) - f(x2))/(x1 - x2) by m .. and x1 and x2 by x

Answer 11

dy/dx = (y2-y1)/(x2-x1)

Answer 12

not possible without calc because "slope of the tannet line" is a calc concept

Answer 13

*tangent

Answer 14

ok @satellite73 can u show it plzz

Answer 15

course i could be wrong

Answer 16

no prob

Answer 17

ah. i have seen methods using circles which work on certain curves though

Answer 18

no i was agreeing with @eigenschmeigen
take the derivative, set it equal to 12 and solve

Answer 19

i'll try and find it

Answer 20

i got x= 7 x =4

Answer 21

is that correct

Answer 22

find the corresponding y

Answer 23

ok

Answer 24

thxxx @experimentX @satellite73 @eigenschmeigen

Answer 25

lol ... yw,
you find the tangent right??

Answer 26

yeah..........