Question

How do you find the zeros, real and imaginary, of `y=1/4(x+4)^2+4` using the quadratic formula?

Answer 1

`color(red)(x=-4+4i, -4-4i)`

Explanation:

Finding the zeros/roots of an equation of the form `y=f(x)` means finding solutions of the equation `f(x)=0`; `i.e.` finding those values of `x` for which `f(x)=0`

`therefore` zeros/roots of `color(red)(y=1/4(x+4)^2+4)` are basically solutions of `color(red)(1/4(x+4)^2+4=0)`

`1/4(x+4)^2+4=0`
`implies [(x+4)^2+16]/4=0`

`implies (x^2+8x+16)+16=0`
`implies x^2+8x+32=0`

Using the quadratic formula;

`x=(-8+-sqrt(8^2-4*1*32))/(2*1)`

`x=(-8+-sqrt(-64))/2 = (-8+-8i)/2 = -4+-4i`

`therefore color(red)(x=-4+4i, -4-4i)` are the two imaginary zeros/roots of the given equation.