How do you find the domain, range, and the zeros of f(x)= x2-12x+36?

Answer 1

first of all what is domain.
it is set of all possible values of x
u can always see that for a quadratic eq. it is always all real no.
range: value of f(x) for valid x which is all real here.
to find range put f(x)=y
u get y=x^2-12x+36
y=(x-6)^2
u can see that f(x) >0 or f(x)=0
hence range is set of whole no.
zeros: value of x for which f(x)=0
hence there is only 1 value that is 6

Answer 2

f(x) = x^2 - 12x + 36

1) Rule of thumb: For polynomials, the domain is ALWAYS the set of real numbers. This includes linear functions, quadratic, cubics, quartics, quintics ....

That means the domain of f(x) is all real numbers. In interval notation, it would be (-infinity, infinity).

2) The range of this function is determined by its vertex. We have to convert this function into the form
f(x) = a(x - h) + k.

(h, k) would represent the coordinates of the vertex, and "a" represents whether we have a minimum vertex or a maximum vertex. If "a" is positive, we have a minimum, and our range is
[k, infinity). If "a" is negative, we have a maximum, and our range is (-infinity, k].

To put this in vertex form, we have to complete the square.

f(x) = x^2 - 12x + 36
f(x) = (x - 6)^2
f(x) = 1(x - 6)^2 + 0

Meaning our vertex is located at (6,0) and our range is
[6, infinity).

3) The zeros are found by equation f(x) to 0.
Since f(x) = x^2 - 12x + 36, we solve for x when

x^2 - 12x + 36 = 0. Factoring this, we get
(x - 6)^2 = 0. This means
x - 6 = 0, and
x = 6.

Therefore, our zeros (or zero, in this case) is 6.