How to use calculus to find the equation of a parabola that is tangent to a quartic

Question

I am stuck on this homework question:

Use calculus to find the equation for a parabola in the form $y = ax^2 + bx + c$ with an $x-$intercept at $(2, 0)$ if it is tangent to the curve $y = (x^2 - 4x)^2$ at $x = 1$.

The answer is $y = -21x^2 + 54x - 24$

Answer 1

you have two equations:

$y = ax^2 + bx + c\\ y = x^4 - 8x^2 + 16$

And you must solve for $a,b,c$ such that the two curves coincide at the same point when $x = 1$ $y(1) = y(1)\\ a+b+c = 9$

Have the same slope at that point

$y'(1) = y'(1)\\ 2a + b = ??$

and for the parabola

$y(2) = 0\\ 4a + 2b + c = 0$

Answer 2

Hints:

You need three pieces of information to find the values of $a$, $b$, and $c$.

First, you know that the parabola goes through the point $(2,0)$. Second, you know that the parabola goes through the same point that the other curve does at $x=1$. Third, you know that the parabola has the same slope that the other curve does at $x=1$.

Use each datum to find an equation in $a$, $b$, and $c$. Then solve those simultaneous equations.