Posted by **Avik** on Sunday, February 3, 2013 at 3:17pm.

Show that the tangent to the curve 25x^5+5x^4-45x^3-5x^2+2x+6y-24=0 at the point (-1,1) is also a normal at two points of the curve.

- Math -
**Reiny**, Sunday, February 3, 2013 at 5:03pm
whewww!

125x^4 + 20x^3 - 135x^2 - 10x + 2 + 6dy/dx = 0

at (-1,1)

125 - 20 - 135 + 10 - 2 + 6dy/dx = 0

6dy/dx = 22, so the slope of the tangent at (-1,1) is 22/6 = 11/3

equation of tangent:

1= (11/3)(-1) + b

b = 1 + 11/3 = 14/3

y = (11/3)x + 14/3 = (11x+14)/3

sub that into the original equation

25x^5+5x^4-45x^3-5x^2+2x+6y-24=0

25x^5+5x^4-45x^3-5x^2+2x+6(11x+14)/3-24=0

25x^5+5x^4-45x^3-5x^2 + 24x + 4=0

after dividing out the factor (x+1) we are left with

25x^4 - 20x^3 - 25x^2 +20x+4=0

tried to solve this, even had Wolfram page try it

http://www.wolframalpha.com/input/?i=25x%5E4+-+20x%5E3+-+25x%5E2+%2B20x%2B4%3D0

it gave me x = appr -.9507 and appr -.169872

If we sub those values into the derivative equation we should get -3/11 or -.2727..

I did not get that.

Can't find my error

OR,

those two points don't exist

(you might print out my solution and go through it to see if find any errors)

- Math -
**Steve**, Sunday, February 3, 2013 at 5:12pm
-6y = 25x^5+5x^4-45x^3-5x^2+2x+24

y' = -1/6 (125x^4 + 20x^3 - 135x^2 - 10x + 2)

y'(-1) = 3

normal has slope -1/3

Go to http://rechneronline.de/function-graphs/

and plot

-1/6 * (25x^5+5x^4-45x^3-5x^2+2x-24)

and its derivative

as well as

3x+4

The line is not normal to the curve anywhere. Typo somewhere here?

- Math - PS -
**Steve**, Sunday, February 3, 2013 at 5:43pm
When plotting the graphs, set

x: -2 to 2

y: -10 to 10

that will show things quite clearly.

## Answer this Question

## Related Questions

- Business Statistics - In each case, sketch the two speciﬁed normal curves...
- Calculus - Consider the curve y^2+xy+x^2=15. What is dy/dx? Find the two points ...
- Calculus - Consider the curve given by y^2 = 2+xy (a) show that dy/dx= y/(2y-x...
- 12th Grade Calculus - 1. a.) Find an equation for the line perpendicular to the ...
- Calculus - Please help this is due tomorrow and I dont know how to Ive missed a ...
- calculus - Consider the curve defined by 2y^3+6X^2(y)- 12x^2 +6y=1 . a. Show ...
- calculus - 1. Given the curve a. Find an expression for the slope of the curve ...
- math - consider the curve defined by the equation y=a(x^2)+bx+c. Take a point(h,...
- calculus!URGENT - Show that the tangent line to the curve y=x^3 at the point x=a...
- Math- Calc 1 - 1. At what point does the normal line to the curve x^2 - XY + Y^2...