Sketch the line y=x^2-4x+3. Be sure to clearly show all the points where the line crosses the coordinate axis and the stationary points

From the equation we can see the the line in a positive quadratic graph. In order to find the points where the line crosses the x axis we must let y=0 and solve for x. We can then use either inspection, completing the square, or the quadratic formula to factorise the quadratic present. By doing this we get (x-1)(x-3)=0. Therfore we can solve to get 1 and 3 when y=0. Therfore the line crosses the x axis at the points (0,0) (1,0) and (3,0). We can also determine that the lilne crosses the y axis at the point (0,3).

In order to find the stationary point of the line we must differentaite with respect to x to get dy/dx=2x-4. When letting dy/dx=0 we get x=2. We can find the y coordinate of the stationary point by substituting x=2 into the original equation to get y=-1. We know that this is a minimum point as it is a postive quadratic. from this we can draw the graph.