(i) Find the point(s) where the curve y=x^2-2x+1 crosses the x-axis, and (ii) find the coordinates of the vertex of the curve.
(i) Recall that a curve crosses the x-axis (the horizontal axis) when the vertical y-coordinate is zero, so we are required to find all the points such that y=0, that is we must solve the equation x2-2x+1=0. Firstly, we should try to factorise the left-hand side, as if this is possible this is normally quicker than attempting to complete the sqaure or use the quadratic formula. To factorise, some people, with practise, find it easiest to factorise 'by inspection', that is just try out different factorisiations quickly in their head until one works. To do this more slowly, we're looking for some numbers a and b such that (x+a)(x+b)=x2-2x+1. Expanding the left-hand side, we obtain x2+ax+bx+ab=x2+(a+b)x+ab=x2-2x+1. So we want a and b such that a+b=-2 and ab=1. Let's try something easy to work out that looks like it might work. Say a=-1, b=-1. Then, a+b=-2 and ab=1. It works! So, x2-2x+1=(x-1)(x-1)=(x-1)2. Now, we must solve (x-1)2=0. Square rooting both side we obtain x-1=0 so x=1. So when x=1, y=0, so the curve y=x2-2x+1 crosses the x-axis at (1,0), as required. (ii) Now, recall that the vertex is the point when the curve is at its least or greatest, depending on if it's a 'happy' parabola (with x2 in the equation) or a 'sad' parabola (with -x2 in the equation). Ours is a happy parabola, so we're looking for the minimum value of x2-2x+1. To do this we need to complete the square, which we do by halving the x coefficient (the number in front of the x, in this case -2), adding this to x and squaring this all, and taking away the square of half the x coefficient. This sounds really confusing! But, it's much easier to understand if you see it done or have done it before. So, we can write x2-2x+1=(x+(-2)/2)2-((-2)/2))2+1=(x-1)2-(-1)2+1=(x-1)2-1+1=(x-1)2. We now have this alternative form for the equation of our curve which makes it easier to find the minimum value and the x-value for which this minimum is obtained (at the vertex). Note that this new expression (x-1)2 is a square, so is always positive or zero. But, since we want to find the minimum value at the vertex, we want to select the smallest value that (x-1)2 can ever be. Since this sqaure can only be positive or zero, the minimum value of (x-1)2 is zero! This occurs when (x-1)2 =0, rooting both sides gives us x-1=0 and then we can add 1 to both sides to get x=1. And the value of the curve at this point is y=0, so the coordinates of the vertex are (1,0). So in fact, in this case the vertex is where the curve crosses the x-axis, although technically in this case it only touches the x-axis.
