Question

1.
Find a and b so that the graph of y = ax^2+bx+3 has a relative minimum at (2,1)

Answer 1

`a=0.5`

`b=-2`

Explanation:

we have

`y=ax^(2)+bx+3`

For x=2, y=1

substitute

`1=a(2)^(2)+b(2)+3`

`4a+2b=-2` -----> equation A

Remember that

If the function has a relative minimum in (2,1), then the first derivative of the function must be equal to zero when x=2

The first derivative is equal to

`(dy)/(dx)=2ax+b`

so

`0=2a(2)+b`

`b=-4a` ----> equation B

Solve the system of equations A and B by substitution

Substitute equation B in equation A

`4a+2(-4a)=-2`

solve for b

`4a-8a=-2`

`-4a=-2`

`a=0.5`

Find the value of b

`b=-4a` ---->
`b=-4(0.5)=-2`

The quadratic equation is

`y=0.5x^(2)-2x+3`