Answer :
To find the equation of the line tangent to the graph of [tex]\( f(x) = -2x^3 \)[/tex] at the point [tex]\((-1, 2)\)[/tex], we will follow a step-by-step procedure involving the calculation of the derivative, the slope at the point of tangency, and then using the point-slope form of the equation of a line.
1. Determine the function and the point of tangency:
- The function is [tex]\( f(x) = -2x^3 \)[/tex].
- The point of tangency is [tex]\((-1, 2)\)[/tex].
2. Find the derivative of [tex]\( f(x) \)[/tex]:
- To find the slope of the tangent line, we need the derivative of [tex]\( f(x) \)[/tex]. The derivative [tex]\( f'(x) \)[/tex] is given by:
[tex]\[ f'(x) = \frac{d}{dx} (-2x^3) = -6x^2 \][/tex]
3. Evaluate the derivative at the point of tangency (i.e., find the slope):
- The slope of the tangent line at [tex]\((-1, 2)\)[/tex] is given by [tex]\( f'(-1) \)[/tex]:
[tex]\[ f'(-1) = -6(-1)^2 = -6(-1) = -6 \][/tex]
- Therefore, the slope of the tangent line is [tex]\(-6\)[/tex].
4. Use the point-slope form of the equation of a line to find the tangent line:
- The point-slope form of the equation of a line 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 of tangency.
- Substituting the values [tex]\( x_1 = -1 \)[/tex], [tex]\( y_1 = 2 \)[/tex], and [tex]\( m = -6 \)[/tex] into the equation, we have:
[tex]\[ y - 2 = -6(x - (-1)) \][/tex]
- Simplifying the equation:
[tex]\[ y - 2 = -6(x + 1) \][/tex]
[tex]\[ y - 2 = -6x - 6 \][/tex]
[tex]\[ y = -6x - 6 + 2 \][/tex]
[tex]\[ y = -6x - 4 \][/tex]
Therefore, the equation of the tangent line at the point [tex]\((-1, 2)\)[/tex] is:
[tex]\[ y = -6x - 4 \][/tex]
So, in the question formatting style:
The equation of the tangent line is [tex]\( y = -6x - 4 \)[/tex].
1. Determine the function and the point of tangency:
- The function is [tex]\( f(x) = -2x^3 \)[/tex].
- The point of tangency is [tex]\((-1, 2)\)[/tex].
2. Find the derivative of [tex]\( f(x) \)[/tex]:
- To find the slope of the tangent line, we need the derivative of [tex]\( f(x) \)[/tex]. The derivative [tex]\( f'(x) \)[/tex] is given by:
[tex]\[ f'(x) = \frac{d}{dx} (-2x^3) = -6x^2 \][/tex]
3. Evaluate the derivative at the point of tangency (i.e., find the slope):
- The slope of the tangent line at [tex]\((-1, 2)\)[/tex] is given by [tex]\( f'(-1) \)[/tex]:
[tex]\[ f'(-1) = -6(-1)^2 = -6(-1) = -6 \][/tex]
- Therefore, the slope of the tangent line is [tex]\(-6\)[/tex].
4. Use the point-slope form of the equation of a line to find the tangent line:
- The point-slope form of the equation of a line 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 of tangency.
- Substituting the values [tex]\( x_1 = -1 \)[/tex], [tex]\( y_1 = 2 \)[/tex], and [tex]\( m = -6 \)[/tex] into the equation, we have:
[tex]\[ y - 2 = -6(x - (-1)) \][/tex]
- Simplifying the equation:
[tex]\[ y - 2 = -6(x + 1) \][/tex]
[tex]\[ y - 2 = -6x - 6 \][/tex]
[tex]\[ y = -6x - 6 + 2 \][/tex]
[tex]\[ y = -6x - 4 \][/tex]
Therefore, the equation of the tangent line at the point [tex]\((-1, 2)\)[/tex] is:
[tex]\[ y = -6x - 4 \][/tex]
So, in the question formatting style:
The equation of the tangent line is [tex]\( y = -6x - 4 \)[/tex].
