We need to find the x value for which the slope of the tangent line to y = ax^2 + bx + c is 1 at the point on the parabola (0,1). Realize that y = x -1 is the actual equation of the tangent line. From this we see that the slope of the tangent line is m = 1.
Differentiate y = ax^2 + bx + c with respect to x: dy/dx = 2ax + b. This is the slope of the tangent line to the parabola at (0,1) But we already know that the line y = x - 1 is tangent to the parabola at (0,1). Thus, 2ax + b = 1 (because the slope of y = x - 1) is 1). At (0,1), 2a(0) + b = 1, so b=1.
Thus, the equation of the parabola becomes more evident:
a) since (0,1) satisfies this equation, 1 = a(0)^2 + 1(0) + c, so c=1
Then y = ax^2 + 1x + 1, and we need only find the value of a.
Since the graph of y = ax^2 + 1x + 1 passes thru (0,1),
y = 1 = a(0)^2 + 1(0) + 1, or 1 = a(0)^2 + 1. 1 = 1, yes, but this does not tell us the value of a!
Is it possible that a could have any value?
Let a = 1. Then y = 1x^2 + 1x + 1. What is the slope of the tan. line to this curve? Differentiating, dy/dx = 2x + 1. At x = 0, the slope of the tan. line is 1. This agrees with the given y= 1x - 1.
Thus, for a = 1 at least, y = 1x^2 + 1x + 1 is the equation of the parabola.
Graphing this equation and the equation of the given line y = x -1 shows that y = x - 1 is tangent to the graph of y = x^2 + x + 1.
