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

2 Answers
May 23, 2018

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

May 23, 2018

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

Explanation:

f(x)=ax^2+bx , xinRR

  • 1inRR
  • 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