Find an equation of the parabola y = ax2 + bx + c that passes through (0, 1) and is tangent to the line y = 4x − 4 at (1, 0).

We are given the following:
          - parabola passes to both (1,0) and (0,1) 
          - slope at x = 1 is 4 from the equation of the tangent line 

First, we figure out the value of c or the y intercept, we use the second point (0, 1) and substitute to the equation of the parabola. When x = 0, y = 1. So, c should be equal to 1. The parabola is y = ax^2 + bx + 1 

Now, we can substitute the point (1,0) into the equation,
0 = a(1)^2 + b(1) + 1 
0 = a + b + 1
a + b = -1 

The slope at x = 1 is equal to 4 which is equal to the first derivative of the equation.
We take the derivative of the  equation ,
y = ax^2 + bx + 1
y' = 2ax + b 

x = 1, y' = 2
4 = 2a(1) + b 
4 = 2a + b 

So, we have two equations and  two unknowns, 
2a + b = 4 
a + b = -1 

Solving simultaneously,
a = 5 
b = -6

Therefore, the eqution of the parabola is y = 5x^2 - 6x + 1 .