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

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Answer 1 · 2018-05-23T14:40:36

#a = 2# and #b=-4#

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 · 2018-05-23T17:51:52

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

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

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#