Question

The graph of `y=ax^2+bx` has an extremum at `(1,-2)`. Find the values of a and b?

Answer 1

`a = 2` and `b=-4`

Explanation:

Given: `y=ax^2+bx, y(1) = -2`

From the given can substitute 1 for x and 2 for y and write the following equation:

`-2 = a+b" [1]"`

We can write the second equation using that the first derivative is 0 when `x =1`

`dy/dx= 2ax+b`

`0= 2a+b" [2]"`

Subtract equation [1] from equation [2]:

`0 - -2=2a+b - (a+b)`

`2 = a`

`a=2`

Find the value of b by substituting `a = 2` into equation [1]:

`-2 = 2+b`

`-4 = b`

`b = -4`

Answer 2

`f(x)=2x^2-4x`

Explanation:

`f(x)=ax^2+bx` , `x` `in` `RR`

• `1` `in` `RR`
• `f` is differentiable at `x_0=1`
• `f` has an extremum at `x_0=1`

According to Fermat's Theorem `f'(1)=0`

but `f'(x)=2ax+b`

`f'(1)=0` `<=>` `2a+b=0` `<=>` `b=-2a`

`f(1)=-2` `<=>` `a+b=-2` `<=>` `a=-2-b`

So `b=-2(-2-b)` `<=>` `b=4+2b` `<=>`

`b=-4`

and `a=-2+4=2`

so `f(x)=2x^2-4x`