Question

find the equation of the normal to y=x^4-4x^3 at the point for which x=1/2.

Answer 1

Normal means perpendicular, so find the slope of the tangent line at that point, then find the perpendicular slope.

Answer 2

You'll also need the y-coordinate of that point too, so evaluate the function at x=1/2 to find y.

Answer 3

Are you comfortable with taking derivatives, @emzy_777 ?

Answer 4

so would i have to do f(1/2)=4x^3-12x^2?

Answer 5

\(f '(x)=4x^3-12x^2\)
Find f '(1/2) to get the slope at that point.

Answer 6

Also find f(1/2) to get the y-coordinate at that point - you'll need that to write the equation of the line after you find the perpendicular slope.

Answer 7

Getting it, @emzy_777 ?

Remember that the slope of a perpendicular line is the opposite reciprocal of the slope of the tangent line.

Answer 8

i got f'(1/2)=-2 1/2 ... i dont think im doing this correctly...

Answer 9

That looks fine so far.
So that -5/2 is the slope of the tangent line. What is the slope of the line perpendicular to that?

Answer 10

2/5?

Answer 11

?

Answer 12

Right. Now you have the slope of the line and a point the line passes through, now it's an Algebra-1 problem to get the equation of that line.

Answer 13

im not sure exactly how to do that

Answer 14

You want the equation of a line. The slope of that line is 2/5, and it passes through the point (1/2, -7/16) - You did already find that point, right?

You can use slope-intercept form or point-slope form to put that information into a linear equation.

Answer 15

i didnt find the point

Answer 16

how do you find the point??

Answer 17

?

Answer 18

Sorry, computer froze on me...

Like I said in my second reply, "You'll also need the y-coordinate of that point too, so evaluate the function at x=1/2 to find y."