Finding the Horizontal Tangent Line?

Question

I'm supposed to find the horizontal tangent line of y=x^3+3x^2+x+3. I know I should find the derivative, set it equal to zero, and then factor. However, the derivative doesn't factor, so I don't know what to do. Any help?

Answer 1

You could try solving the quadratic equation that you get from setting the first derivative to equal zero by using the formula.

Edited to say that the discriminant is greater than zero, so there is a solution, but it's not rational.

Answer 2

y = (x^3) + 3(x^2) + x + 3

dy/dx = 3(x^2) + 6x + 1 = 0

Now, as you correctly said, the derivative doesn't factor. But there is a clue in the question: the word "horizontal". This means that the gradient of the tangent line is zero. Therefore:

d(dy/dx)dx = 6x + 6 = 0

Therefore, x = -1

Substitute in your original equation to get your y co-ordinate. And carry on from there using y = mx + c. Since it is a horizontal line, your m = 0, so you'll get y = c.

Good luck!

Answer 3

Hi,

Answer:  y = 5.0886 and y=2.911

How to do it:

  You’ll remember that the second derivative is zero at the inflection point, not at a relative max or min when the slope of the tangent line is zero.  

So, we have to solve the nasty quadratic equation.

X=[-6 ±√(6²-4(3)1)]/(2*3)

X=[ -6 ±√(24)]/6

X=-1±0.81649

X=-.184

X=1.816

So, what you do next is put these two numbers in the cubic equation and solve for y.  You can do this on just about any scientific or graphing calculator.  (Yes, you can do it on *most* scientifics.) You’ll then have y= 5.0886 and y=2.911, or maybe something very close to that.

Sorry, but I got lazy and used a graphing calculator for those values.  Besides, I didn’t want to deprive you of the excitement of cranking out those values in the cubic equation. :-).  

Hope this helps.  

Incidentally, if you have a graphing calculator, want to use it, and don’t know how to this one, you can get some help on my website at this URL:  www.angelfire.com/pro/fkizer

FE