How do I find the coordinates of the turning point of the graph: y=3x^2-6x+7 ?

Our first step here is to differentiate the equation of the graph. To do this we will differentiate each of the three parts in the equation, as follows: To differentiate 3x^2, we first multpily the 3 in front of the "x" by the power, which in this case is 2, giving us an answer of 6. The 6 will then go before our "x" term. We then lower the power by 1, leaving us with a final differentiated term of 6x. To differenitate the term "6x", we must first recognise that "6x" is the same as "6x^1". We therefore multiply the 6 in front of the x by 1, giving us 6, and lower the power by 1, giving us zero. Our final term is 6x^0, which is the same as 6. When we differentiate the 7, we simply get zero. This is because 7 is the same as 7x^0. We know that if we multiply any number by zero we get an answer of zero, therefore our final differentiated term is zero.Our differentiated equation, putting all the terms together is now: dy/dx= 6x-6At any turning point, the gradient of the graph must be zero. Therefore, to find our "x" coordinate of the turning point we write out the equation: 6x-6=0. If we add 6 to both sides of the equation, it gives us 6x=6. We can then divide both sides by 6, giving us x=1. We now know that the x coordinate of our turning point is 1. To find the y coordinate, we simply plug the 1 into the original equation, replacing "x" with 1. Therefore: y= (31^2)-(61)+7= 4. The coordinates of our turning point are therefore: (1,4)