Find the coordinates of the turning point of the curve y=x^2+3x+7

We know that turning points occur when the gradient is equal to zero. Hence, we differentiate this curve. dy/dx = 2x+3 and we set this equal to zero. This gives 2x+3=0, we then rearrange to get 2x=-3 and so x=-3/2. Placing this value of x back into the curve equation gives y=(-3/2)^2+3(-3/2)+7 and so y= 19/4. Therefore the coordinates of the turning point are (-3/2,19/4)