Calculus Question

Question

Find an equation of the parabola y = ax² + bx + c that passes through (0, 1) and is tangent to the line y = 5x − 5 at (1, 0).

Answer 1

Henry:

If y = ax² + bx + c passes through (0,1), then 1 = a*0² + b*0 + c or c = 1

So we know y = ax² + bx + 1 (where I sub c=1)

If the parabola is tangent to the line y = 5x-5 at (1,0), then the parabola must have the same slope (which is 5) at point (1,0)

The slope of the parabola is given by its derivative: y' = 2ax + b

We need the derivative to equal 5 at point (1,0), so 5 = 2*a*(1) + b or 5 = 2a+b

We also know that the parabola passes through point (1,0) - since the parabola is tangent there- so

0 = a*1² + b*1 + 1 or a+b = -1

We can now solve for a and b using the simultaneous equations 2*a + b = 5 and a + b = -1

Let me know if you need more help!