To find the equation(s) of the tangent line(s) to the graph of the curve [tex]\( y = x^3 - 12x \)[/tex] passing through the point [tex]\( (1, -12) \)[/tex], we follow these steps:
1. Identify the curve and the given point:
- Curve: [tex]\( y = x^3 - 12x \)[/tex]
- Point: [tex]\( (1, -12) \)[/tex]
2. Differentiate the curve to find the slope of the tangent line:
- The derivative of [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex] is [tex]\( \frac{dy}{dx} \)[/tex].
- [tex]\( y = x^3 - 12x \)[/tex]
- Differentiating: [tex]\( \frac{dy}{dx} = 3x^2 - 12 \)[/tex]
3. Evaluate the derivative at the point of tangency to find the slope:
- Substitute [tex]\( x = 1 \)[/tex] into [tex]\( \frac{dy}{dx} \)[/tex]:
- [tex]\( \frac{dy}{dx} = 3(1)^2 - 12 = 3 - 12 = -9 \)[/tex]
4. Use the point-slope form of a line to find the equation of the tangent line:
- The point-slope form is [tex]\( y - y_1 = m(x - x_1) \)[/tex], where [tex]\( m \)[/tex] is the slope, and [tex]\( (x_1, y_1) \)[/tex] is the point.
- Here, [tex]\( m = -9 \)[/tex] and the point is [tex]\( (1, -12) \)[/tex].
- Substituting these values:
[tex]\[
y - (-12) = -9(x - 1)
\][/tex]
Simplifying the equation:
[tex]\[
y + 12 = -9x + 9
\][/tex]
[tex]\[
y = -9x + 9 - 12
\][/tex]
[tex]\[
y = -9x - 3
\][/tex]
Therefore, the equation of the tangent line to the graph of the curve [tex]\( y = x^3 - 12x \)[/tex] that passes through the point [tex]\( (1, -12) \)[/tex] is:
[tex]\[ y = -9x - 3 \][/tex]
