Complete the square and hence sketch the graph of f(x) = x^2 + 2x + 7

Consider (x + 1)^2, where we halved the coefficient of x to obtain 1. This gives us x^2 + 2x + 1, which looks similar to our function, except it differs by a constant value of 6, hence, by adding 6 to (x + 1)^2, we get f(x)= (x+1)^2 + 6, so we have completed the square. This format of f(x) makes the graph far easier to sketch, as now we know that (x+1)^2 >= 0 for every value of x, so the minimum value of f(x) is 6, where (x + 1)^2 = 0, which means that (-1,6) is the minimum point of f(x). Since the graph never crosses at y = 0, there are no solutions to f(x) = 0, so the only axis we need to consider the graph crossing is now the y-axis. Where x = 0, f(x) = (1)^2 + 6 = 7, so the graph has the 'usual' quadratic shape with y- intercept (0,7) and min point at (-1,6).