Question

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?

Answer 1

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`