How do you write a quadratic function in standard form whose graph passes through points #(1/2, -3/16)#, #(3,-17/4)#, #(4,-17/2)#?

1 Answer
George C.
Feb 22, 2018

#f(x) = -3/4x^2+x-1/2#

Explanation:

Note that we can define a quadratic that takes the value #1# when #x=1/2# and the value #0# when #x=3# or #x=4# by writing:

#((x-3)(x-4))/((1/2-3)(1/2-4)) = ((x-3)(x-4))/((-5/2)(-7/2)) = 4/35(x-3)(x-4)#

Similarly, we can define a quadratic that takes the value #1# when #x=3# and the value #0# when #x=1/2# or #x=4# by writing:

#((2x-1)(x-4))/((6-1)(3-4)) = -1/5(2x-1)(x-4)#

Similarly, we can define a quadratic that takes the value #1# when #x=4# and the value #0# when #x=1/2# or #x=3# by writing:

#((2x-1)(x-3))/((8-1)(4-3)) = 1/7(2x-1)(x-3)#

We can add suitable multiples of these quadratics to get one with the desired properties:

#f(x) = 4/35(x-3)(x-4)(-3/16) - 1/5(2x-1)(x-4)(-17/4) + 1/7(2x-1)(x-3)(-17/2)#

#color(white)(f(x)) = -3/140(x-3)(x-4) + 17/20(2x-1)(x-4) - 17/14(2x-1)(x-3)#

#color(white)(f(x)) = 1/140(-3(x-3)(x-4) + 119(2x-1)(x-4) - 170(2x-1)(x-3))#

#color(white)(f(x)) = 1/140(-3(x^2-7x+12) + 119(2x^2-9x+4) - 170(2x^2-7x+3))#

#color(white)(f(x)) = 1/140(-3x^2+21x-36 + 238x^2-1071x+476 - 340x^2+1190x-510)#

#color(white)(f(x)) = 1/140(-105x^2+140x-70)#

#color(white)(f(x)) = 1/4(-3x^2+4x-2)#

#color(white)(f(x)) = -3/4x^2+x-1/2#

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