Two lines are perpendicular iff (if and only if) the product of their gradients is -1. Therefore to work out the gradient of a normal you have to find the gradient of the tangent. The gradient of the normal will be −1gradientoftangent

The gradient of a tangent at the point (x,y(x)) is the derivative of y(x) at x.

let f(x) be the gradient of the normal at the point x.

f(x)=−1ddx(2×2–3x)=−14x−3

Edit: Has someone changed the question, because when I answered the question it was “what is the gradient of the normals of the curve 2x^2–3x” (or something similar to

## Answers ( )

Answer:Hope this helps !!! 🙂

Step-by-step explanation:Two lines are perpendicular iff (if and only if) the product of their gradients is -1. Therefore to work out the gradient of a normal you have to find the gradient of the tangent. The gradient of the normal will be −1gradientoftangent

The gradient of a tangent at the point (x,y(x)) is the derivative of y(x) at x.

let f(x) be the gradient of the normal at the point x.

f(x)=−1ddx(2×2–3x)=−14x−3

Edit: Has someone changed the question, because when I answered the question it was “what is the gradient of the normals of the curve 2x^2–3x” (or something similar to