we are given function as
[tex] f(x)=3x^2-12x+5 [/tex]
Axis of symmetry:
we can use formula to find axis of symmetry
[tex] f(x)=ax^2+bx+c [/tex]
[tex] x=-\frac{b}{2a} [/tex]
a=3 , b=-12
[tex] x=-\frac{-12}{2*3} [/tex]
now, we can solve for x
[tex] x=2 [/tex]
so, axis of symmetry is [tex] x=2 [/tex].........Answer
Domain:
we know that
domain is all possible values of x for which any function is defined
since, it is quadratic function
so, it is defined for all real values of x
so, we get
[tex] (-\infty,\infty) [/tex]
Range:
we know that
range is all possible value of y
we can plug vertex x=2 into f(x) and find y
[tex] f(2)=3(2)^2-12(2)+5 [/tex]
[tex] y=-7 [/tex]
Since, we have leading coefficient is 3
so, parabola will be open upward
so, smallest y-value will be -7
so, range will be
[tex] [-7,\infty) [/tex]
