tangent line problem

y = ax^2 + b        y  = 2x^2 + cx

Since the point (-1, 0)   is on both graphs

Using the first function, we have

0 = a(-1)^2 + b

0  =  a  + b

Using the second function, we have that

0 = 2(-1)^2 + c(-1)

0 = 2  - c

c = 2

Taking the derivative of both functions

y'  = 2ax               y'  = 4x + c

y' = 2ax               y' =  4x + 2

The slope of the tangent line is the same for both functions at x  = -1

So....equating slopes

2a(-1)   =  4(-1) + 2

-2a  =    -4 + 2

-2a  = = -2

a  =  1

And since a + b  = 0

b  =  -1

The equation of the tangent line is

y =  (4(-1) + 2) ( x  - -1)

y =  -2(x + 1)

y =  -2x -2

Here's a graph  :  https://www.desmos.com/calculator/2wiesd0ekz