# Thread: Horizontal Tangent & Point of Inflection

1. ## Horizontal Tangent & Point of Inflection

the function y=x^4+bx^2+8x+1 has a horizontal tangent and a point of inflection for the same value of x. What must be the value of b?

To do this i know that y'(x)=y''(x) because they both equal 0 but i tried that but I don't know how to solve because i don't know x.

2. $y=x^4+bx^2+8x+1$

$y'=4x^3+2bx+8$

$y''=12x^2+2b$

Set y' and y'' equal to zero, since a horizontal tangent has slope = 0 and the point of inflection is where second derivative=0 (changes curvature/concavity).

You now have a system of 2 equations. Solve for b and x.

3. Ok, so your issue is with solving the system of equations. Two equations. Two unknowns.

$4x^3+2bx+8 = 2(2x^3 + bx + 4) =0 \implies 2x^3 + bx + 4 = 0$

$12x^2+2b = 2(6x^2 + 1)=0 \implies 6x^2 + b= 0$

Does this help?

4. not really

5. $y=x^4+bx^2+8x+1$

$y'=4x^3+2bx+8$

$y''=12x^2+2b$

Set y' and y'' each = to 0.

$4x^3+2bx+8 = 2(2x^3 + bx + 4) =0 \implies 2x^3 + bx + 4 = 0$

$12x^2+2b = 2(6x^2 + 1)=0 \implies 6x^2 + b= 0$

$
6x^2 + b = 0 \implies b = -6x^2
$

$
\implies 2x^3 + bx + 4 = 2x^3 + (-6x^2)x +4 = 0
$

Any better now? You can solve for x now, without the b term. All I did was substitution.

6. I get it now! Thank you so much!