Find the equation of the parabola y = ax^2 + bx + c that is tangent to the line y = x - 1 at (0,1)

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.

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