Find the 1st derivative of y = x^2 + 7x +3 and hence find the curves minima.
Firstly, we differentiate y = x2+7x+3 . This gives dy/dx = 2x+7.The minimum value occurs when dy/dx = 0. So find x and y when dy/dx=0. 2x+7=0 implies x= -3.5, which from the first equation means y = (-3.5)2 + 7*3.5 +3 = 39.75.Therefore, the minimum value has the position (-3.5, 39.75).
CW
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