Find the value of a,b,c for which the parabola y=ax^2+bx+c

Question

passes through point(1,3) and has the tangent y=-4x+9 at point (2,1)?

Answer 1

The parabola equation is  y=ax^2+bx+c .

The point (1,3) passes through parabola so it satisfy the curve .

3 = a (1²) + b(1) +c

3 = a + b + c   --------->(1)

The tangent point will also satisfy the parabola .

So (2,1) will satisfy the curve .

1 = a (2²) + b(2) +c

1 = 4a + 2b + c    -----------------> (2)

To find out the tangent , equate the first derivative at (2,1) .

y' = 2ax + b

y' = 2a(2) + b

y' = 4a + b

The given tangent is y = -4x+9 .

Slope of tangent is -4 .

So  4a + b = -4    -----------------------> (3)

Now solve the above 3 equations .

Subtract equation (2) from (1) 

4a + 2b + c = 1

 -a -   b -  c = -3

----------------------

3a + b = -2  -------------------->(4)

Subtract equation (3) from (4)

4a + b = -4

-3a - b = +2

-----------------

a + 0 = -2

a = -2

Put a = -2 in equation (3)

4(-2) + b = -4

-8 + b = -4

b = 4

Put b =4 and a = -2 in  equation (1)

3 = -2 + 4 + c 

3 = 2 +c

c = 3 - 2 

c = 1

So the values are  a = -2 , b =4 and c = 1 .