Final answer:
The gradient of the normal to the curve y = x^2 - 7x - 4 at the point (1,0) is calculated by first finding the derivative of the curve's equation and then taking the negative reciprocal of the derivative at x = 1, which results in 1/5.
Step-by-step explanation:
To find the gradient of the normal to the curve y = x^2 - 7x - 4 at the point (1,0), we must first calculate the gradient of the tangent to the curve at that point. The gradient of the tangent is the derivative of the curve's equation evaluated at x = 1.
The derivative of y = x^2 - 7x - 4 with respect to x is dy/dx = 2x - 7. Substituting x = 1 gives us dy/dx = 2(1) - 7 = -5, which is the gradient of the tangent at the point (1,0).
The gradient of the normal is the negative reciprocal of the gradient of the tangent. Therefore, the gradient of the normal to the curve at the point (1,0) is -1/(-5) = 1/5.