Question

Consider the curve:
y=x^3-(3/2)x^2-6x+10

Find the points and equations for the tangent where the tangent is perpendicular to
the line:

y=1-(x/12)

Answer 1

Helping need please

Answer 2

when two lines are perpendicular, the product of their slopes equals -1. are you you familiar with the slope-intercept form of a line?

Answer 3

no..

Answer 4

can you show the solution..so I can detect my problem

Answer 5

y = mx + c, here m is the slope and c is the intercept made by the line on the y axis. rewrite the given equation in this form, you'd get
y = (-1/2) x + 1. SO the slope is m = (-1/2). using m m' = -1
we can find the value of m'.

Answer 6

oh..ok..I try...thanks

Answer 7

we get, m' = 2. slope of a tangent can be found by the derivative of the curve. so find dy/dx and put its value equal to 2.

Answer 8

\[\frac{dy}{dx} =3x^2 - 3x - 6\] this gives us the equation \[ 2 =3x^2 - 3x - 6\] \[3x^2-3x -8 = 0\] this is a quadratic equation, solve to find its roots. you'd get two values for x. put these values in the equation of the curve and find the two corresponding values of y.

Answer 9

i made a mistake in reading the question the slope of the given line is -1/12, so m' = 12. thus the quadratic equation would become 3x^2-3x-6 = 12
3x^2-3x-18 = 0
3(x^2-x-6) = 0
(x-3)(x+2) = 0
x = 3 and x = -2

Answer 10

how you get the equation 3x^2-3x-6=12?

Answer 11

by differentiating the given equation of the curve, y=x^3-(3/2)x^2-6x+10 and putting it equal to the slope found by us.
The corresponding values of y can be found by putting x = 3 and then then x = -2 inthe equation y=x^3-(3/2)x^2-6x+10.
for example if x = 1, y = (1)^3 -(3/2)(1)^2 -6 (1) +10 = 7/2

Answer 12

ok...I see my solution...thank you

Answer 13

its been a pleasure :) Have a very prosperous new year.