Find equations of the tangent lines at all inflection points of the graph of

f(x)=x^4 - 6x^3 - 12x^2 - 8x + 3

To find the equations of the tangent lines at the inflection points of the function f(x) = x^4 - 6x^3 - 12x^2 - 8x + 3, we first need to find the inflection points.

An inflection point occurs when the concavity of the function changes, which means the second derivative changes sign. So, we'll start by finding the second derivative of f(x).

1. Find the first derivative of f(x):
The first derivative of f(x) is obtained by differentiating each term with respect to x:
f'(x) = 4x^3 - 18x^2 - 24x - 8

2. Find the second derivative of f(x):
Differentiate the first derivative again:
f''(x) = 12x^2 - 36x - 24

3. Set f''(x) equal to zero and solve for x:
12x^2 - 36x - 24 = 0

4. Factor the quadratic equation:
4(3x^2 - 9x - 6) = 0

5. Solve for x using the quadratic formula or factoring techniques:
By factoring 3x^2 - 9x - 6, we obtain:
(3x - 6)(x + 1) = 0

Setting each factor to zero, we find:
3x - 6 = 0, which gives x = 2
x + 1 = 0, which gives x = -1

Therefore, the values of x at the inflection points are x = -1 and x = 2.

6. Plug these values of x into the original function f(x) to get the corresponding y-coordinates:
When x = -1, we have:
f(-1) = (-1)^4 - 6(-1)^3 - 12(-1)^2 - 8(-1) + 3
= 1 + 6 - 12 + 8 + 3
= 6

When x = 2, we have:
f(2) = (2)^4 - 6(2)^3 - 12(2)^2 - 8(2) + 3
= 16 - 48 - 48 - 16 + 3
= -93

Therefore, the inflection points of the graph of f(x) are (-1, 6) and (2, -93).

7. Find the slope of the tangent lines at the inflection points:
The slope of a tangent line at a specific point on a curve is given by the derivative of the function evaluated at that point. So, we need to find the first derivative of f(x), and then evaluate it at x = -1 and x = 2.

Using the first derivative we found earlier:
f'(x) = 4x^3 - 18x^2 - 24x - 8

When x = -1:
f'(-1) = 4(-1)^3 - 18(-1)^2 - 24(-1) - 8
= -4 - 18 + 24 - 8
= -6

When x = 2:
f'(2) = 4(2)^3 - 18(2)^2 - 24(2) - 8
= 32 - 72 - 48 - 8
= -96

Therefore, the slopes of the tangent lines at the inflection points are -6 and -96, respectively.

8. Use the point-slope form to find the equations of the tangent lines:
The point-slope form of a linear equation is given by:
y - y1 = m(x - x1)

Using the slope-intercept form y = mx + b, we can rewrite the point-slope form as:
y = mx + (y1 - mx1)

For the inflection point (-1, 6):
y = -6x + (6 - (-6)(-1))
= -6x + 6 - 6
= -6x

Therefore, the equation of the tangent line at the point (-1, 6) is y = -6x.

For the inflection point (2, -93):
y = -96x + (-93 - (-96)(2))
= -96x - 93 + 192
= -96x + 99

Therefore, the equation of the tangent line at the point (2, -93) is y = -96x + 99.

So, the equations of the tangent lines at the inflection points of the graph of f(x) = x^4 - 6x^3 - 12x^2 - 8x + 3 are y = -6x and y = -96x + 99.