The quadratic function has real coefficients, a zero at 1 + 3i, and a y-intercept of 20. How do I put this in standard form?

1 Answer
Jan 18, 2018

Let the standard form of the quadratic function be

#f(x)=ax^2+bx+c#

The function has a y-intercept of 20,

So #f(0)=20#

#=>axx0^2+bxx0+c=20#

#=>c=20#

The function becomes

#f(x)=ax^2+bx+20#

#f(x)# has got zero at #1+3i#. Hence #f(x)=0# will have two complex roots #alpha=(1+3i)and beta=1-3i#

So sum of the roots
#alpha+beta=-b/a#

#=>(1+3i)+(1-3i)=-b/a#

#=>b/a=-2.......[1]#

Again the product of the roots
#alphaxxbeta=20/a#
#=>(1+3i)xx(1-3i)=20/a#

#=>10=20/a#

#=>a=2......[2]#

Multiplying [1] by [2] we get

#=>b=-4#

So the standard form of the quadratic function becomes

#f(x)=2x^2-4x+20#