 Case 2: One Solution (Line is Vertical) Line: x = 1 (vertical line) Parabola: y = 2x2 - 8x + 11 Graphical Solution: The vertical line doesn't require a table; it's a vertical line at 1. To graph the parabola, this time we'll complete the square first: y = 2x2 - 8x + 11 y = 2(x2 - 4x + 4) + 11 - 8 y= 2(x - 2)2 + 3   The solution seems to be the point (1, 5) By substituting this point into each of the equations to check it, (not shown) we can see that it works. Solution (1, 5) Algebraic Solution: x = 1 y = 2x2 - 8x + 11 Because the intersection point must be on the vertical line, we know its x coordinate must be 1. Substitute this into the other equation to find y: y = 2(1)2 - 8(1) + 11 y = 2 - 8 + 11 y = 5 Solution (1, 5) Now let's look at the case where the line is a tangent to the curve >>> |