Find equations of all tangent lines to the graph of y equals 4 x cubed minus 12 x minus 9 which pass through the point left parenthesis negative 1 comma 3 right parenthesis ​(not on the​ curve).

We are tasked with finding the equations of all tangent lines to the curve \( y = 4x^3 - 12x - 9 \) that pass through the point \( (-1, 3) \), which is not on the curve.

### Step 1: Find the derivative
The first step is to find the derivative of the function \( y = 4x^3 - 12x - 9 \), which represents the slope of the tangent line at any point on the curve. 

The derivative is:
\[
y' = \frac{d}{dx}(4x^3 - 12x - 9) = 12x^2 - 12
\]
Thus, the slope of the tangent line at any point \( x = a \) on the curve is \( y'(a) = 12a^2 - 12 \).

### Step 2: Equation of the tangent line
The equation of a tangent line to the curve at a point \( (a, f(a)) \) has the form:
\[
y - f(a) = m(x - a)
\]
where \( m = 12a^2 - 12 \) is the slope of the tangent line at \( x = a \), and \( f(a) = 4a^3 - 12a - 9 \) is the y-coordinate of the curve at \( x = a \).

Therefore, the equation of the tangent line is:
\[
y - (4a^3 - 12a - 9) = (12a^2 - 12)(x - a)
\]
or
\[
y = (12a^2 - 12)(x - a) + (4a^3 - 12a - 9)
\]

### Step 3: Use the point \( (-1, 3) \)
The tangent lines must pass through the point \( (-1, 3) \). Substituting \( x = -1 \) and \( y = 3 \) into the equation of the tangent line:
\[
3 = (12a^2 - 12)(-1 - a) + (4a^3 - 12a - 9)
\]
Now, we simplify this equation to solve for \( a \).

### Step 4: Solve the equation
Simplify the equation step by step:

\[
3 = (12a^2 - 12)(-1 - a) + (4a^3 - 12a - 9)
\]
Expand \( (12a^2 - 12)(-1 - a) \):
\[
3 = (12a^2 - 12)(-1) + (12a^2 - 12)(-a) + (4a^3 - 12a - 9)
\]
\[
3 = -12a^2 + 12 - 12a^3 + 12a^2 + 4a^3 - 12a - 9
\]
Simplify:
\[
3 = (-12a^3 + 4a^3) + (-12a) + (12 - 9)
\]
\[
3 = -8a^3 - 12a + 3
\]
Subtract 3 from both sides:
\[
0 = -8a^3 - 12a
\]
Factor the equation:
\[
0 = -4a(2a^2 + 3)
\]
This gives us two cases to solve:
1. \( -4a = 0 \) which leads to \( a = 0 \).
2. \( 2a^2 + 3 = 0 \), which has no real solution because \( 2a^2 + 3 = 0 \) implies \( a^2 = -\frac{3}{2} \), and the square of a real number cannot be negative.

### Step 5: Find the tangent line for \( a = 0 \)
For \( a = 0 \), the tangent line equation becomes:
- The slope \( m = 12(0)^2 - 12 = -12 \).
- The point on the curve at \( a = 0 \) is \( (0, -9) \).

Thus, the equation of the tangent line is:
\[
y - (-9) = -12(x - 0)
\]
\[
y + 9 = -12x
\]
\[
y = -12x - 9
\]

### Final Answer
The only tangent line to the curve \( y = 4x^3 - 12x - 9 \) that passes through \( (-1, 3) \) is:
\[
y = -12x - 9
\]

Would you like more details or have any questions?

Here are some related questions to explore:
1. How can we find the normal line to a curve at a point?
2. What is the geometric significance of the derivative in finding tangent lines?
3. Can there be multiple tangent lines through a given point for other curves?
4. How can we find tangent lines to parametric equations?
5. What happens if the point lies on the curve itself?

**Tip:** Always double-check whether your derivative calculations and simplifications are correct when solving for tangent lines.

Find equations of all tangent lines to the graph of y equals 4x^3 - 12x - 9 which pass through the point (-1, 3) (not on the curve).

Learn how to find the equations of all tangent lines to the cubic function y = 4x^3 - 12x - 9 that pass through the point (-1, 3). This problem explores key calculus concepts like derivatives and the tangent line formula, with step-by-step solutions. Suitable for high school calculus students.

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