4. For the curve with y = -2x3 + x - 6, find:
(b.) the gradient at the point (-1, -5)
(c.) the equation of the tangent line at the point (-1, -5).
y = -2x^3 + x - 6
dy/dx = -6x + 1
at (-1,-5), dy/dx = -6(-1) + 1 = 7
so y = -7x + b
at (-1,-5)
-5 = -7(-1) + b
b = -12
equation of tangent is y = -7x - 12
y = -2x^3 + x - 6
dy/dx = -6x + 1
at (-1,-5), dy/dx = -6(-1) + 1 = 7
so y = -7x + b
at (-1,-5)
-5 = -7(-1) + b
b = -12
equation of tangent is y = -7x - 12
To find the gradient at a given point on a curve, we need to calculate the derivative of the curve's equation and then substitute the coordinates of the point into the derivative.
Step 1: Find the derivative of the curve's equation.
The given equation of the curve is y = -2x^3 + x - 6. To find the derivative, we differentiate the equation with respect to x.
dy/dx = d/dx(-2x^3) + d/dx(x) + d/dx(-6)
Differentiating each term, we get:
dy/dx = -6x^2 + 1
Step 2: Substitute the coordinates of the point into the derivative.
The point given is (-1, -5). Substitute x = -1 into the derivative:
dy/dx = -6(-1)^2 + 1
dy/dx = -6(1) + 1
dy/dx = -6 + 1
dy/dx = -5
So the gradient at the point (-1, -5) is -5.
To find the equation of the tangent line at a given point, we use the point-slope form of a linear equation, y - y1 = m(x - x1), where (x1, y1) is the given point and m is the gradient.
Given point: (-1, -5)
Gradient: -5
Substituting the values into the equation, we get:
y - (-5) = -5(x - (-1))
y + 5 = -5(x + 1)
y + 5 = -5x - 5
y = -5x - 10
Therefore, the equation of the tangent line at the point (-1, -5) is y = -5x - 10.