Given:" "t(x)=3(x^2+2x)+5
This is part way towards converting standard equation form to vertex form.
color(blue)("Determine the y-intercept")
The constant of 5 is the same as the constant c in y=ax^2+bx+c the y-intercept is y=c=5
color(blue)(=> y_("intercept")->(x,y)->(0,5))
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color(blue)("Determine the x-intercepts for " x in RR)
First of all lets make sure they exist.
Consider the method of solving using the formula where
x=(-b+-sqrt(b^2-4ac))/(2a)
The part that is sqrt(b^2-4ac) is called the determinant.
From sqrt(b^2-4ac)" ; if "b^2-4ac < 0 then there are no x-intercepts in the 'set' of numbers that are called 'Real'. However there are solutions that are in the 'set' of numbers called 'Complex'.
color(brown)(y=ax^2+bx+c)color(blue)(" "->" "y= 3x^2+6x+5)
color(brown)(=>sqrt(b^2-4ac)color(blue)(" "->" "sqrt(6^2-4(3)(5))
sqrt(36-60) and 36-60<0
color(blue)("Thus "x_("intercept") !in RR)
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Slightly higher level of mathematics
color(blue)("Determine the x-intercepts for " x in CC " (complex numbers)")
Completing the square gives:
Standard form" "y=ax^2+bx+c" "->" "y=3x^2+6x+5
" " ->" "a(x+b/(2a))^2+c+k
t(x)=3(x^2+2x)+5" " ->" "t(x)=3(x+1)^2+5+k
where k=(-1)xxa(b/(2a))^2
k=(-1)xx(3xx1^2) =-3
=> t(x)=3(x+1)^2+2
Set to zero
=> 0 = 3(x+1)^2+2
(x+1)^2=-2/3
x=+-sqrt(-1xx2/3)color(white)(.)-1
x=-1+-sqrt(2/3)color(white)(..)( i)
But sqrt(2/3)=sqrt(6)/3
color(blue)(x=-1+-sqrt(6)/3color(white)(..)i
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