Find the equation of the normal to the curve y = 2x^2 -3x +7 at the point x = 1.
Thinking about the normal, it is a line which cuts the curve at right angles, and so will have the form y = mx + c. Now at the point where x = 1, we know the y value from the curve's equation is y = 2(1)2 -3(1) + 7 = 2 -3 + 7 = 6, so both the curve and this line pass through the point (1,6). The normal cuts the curve at right-angles and so its gradient depends of the gradient of the curve. We can find the curve's gradient by differentiating the curve's function in the standard way using the nxn-1 rule to give dy/dx = 4x -3. So the gradient of the curve at x = 1 is 4(1) - 3 = 1. Now, we must be careful as the question asked for the equation of the normal, whereas the gradient of the curve is the gradient of the tangent. However, as they meet at 90o , we can use the fact that the gradient of the normal = -1/ gradient of the tangent, so here we have -1/ 1 = -1. Bringing it all together and again thinking about y = mx +c, we know that for where x = 1, y = 6 and m = -1, so we are just missing c! We can rearrange for c = y -mx = 6 - (-1)(1) = 7, so the final equation is y = -x + 7.
For most questions asking to find equations of tangents and normals, this same method can be applied; think of the equation of the line's form (y = mx+c), use the curve's equation to find the coordinates of where the line sits on the curve (y,x), find the gradient of the curve at this point by differentiating its equation (m) and use gradient of normal = 1 / gradient of tangent if the question asks for the normal, finally substitute back into y= mx +c to find c. Remember to leave the answer in a sensible form, which may be specified in the question!
