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OpenStudy (sshayer):
because imaginary roots always occur in pairs.
so 2 i is also a zero off(x)
\[x= \pm 2i,\]
hence x+2i and x-2 i are factors of f(x)
\[\left( x+2 \iota \right)\left( x-2 \iota \right)=x^2-(2 \iota)^2=x^2-4 \iota ^2=x^2-4*-1=x^2+4\]
divide f(x) by x^2+4
|dw:1472678197269:dw|
\[x^2-9=x^2-3^2=(x+3)(x-3)\]
x+3=0 gives x=-3
x-3=0 gives x=3
all zeros are \[\pm 2 \iota ,\pm 3\]
OpenStudy (iwanttogotostanford):
@sshayer
OpenStudy (iwanttogotostanford):
@retirEEd
OpenStudy (iwanttogotostanford):
@sweetburger @zepdrix
zepdrix (zepdrix):
\[\large\rm g(x)=-(x - 7)^2+1\]This is a quadratic, written in vertex form.
Do you know where the vertex is located?
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