Point A lies on the curve: y=x^2+5*x+8. The x-coordinate of A is -4. What is the equation of the normal to the curve at A?

First we will find the gradient of the tangent of the curve at A. So first, we differentiate y with respect to x. We get that dy/dx=2x+5. We can plug in x=-4 to find the gradient of the tangent at A. We get that dy/dx=2(-4)+5=-8+5=-3. We need to take the negative reciprocal of this to get the gradient of the normal at A. Therefore the gradient of the normal at A is -1/-3=1/3. The general equation for this normal is now y=1/3x+C where C is a constant. To find C, we need to plug in the values of (x,y) at A. To find y at A, then we just need to let x=-4 in the equation of the curve, that we are given. y=(-4)^2+5(-4)+8=16-20+8=4. A(x,y) = (-4,4). Plug these points back into the equation of the normal to find C and we will find that C=16/3. Our general equation is now: y=1/3x+16/3. Multiply through by 3 to get that the equation of the normal to the curve at A is: 3y=x+16.