Domain:
This function is a polynomial, i.e. a sum of powers of a variables, each with its coefficient. Polynomials are defined for every possible value of the variable, so the domain is the whole real number set: [tex] D = \mathbb{R} [/tex]
Range:
A polynomial of degree 2 represents a parabola. Since the leading coefficients, i.e. the coefficient of the term with highest degree, is positive (in this case, it's 3), the parabola is concave up. It means that it has a minimum, and it's unbounded from above. So, the range is something like [tex] [a,\infty) [/tex]. To find the minimum, let's start with the "standard" parabola [tex] y=x^2 [/tex], and transform it to the one of this exercise. [tex] y=x^2 [/tex] has minimum 0, and thus its range is [tex] [0,\infty) [/tex]. When you multiply it times three, its shape narrows, but the range wont change: [tex] [0,\infty) \to [3\cdot 0,3\cdot \infty) = [0,\infty) [/tex]. Finally, when you subtract 5, you shift everythin down 5 units. This transformation affects the range, since you have [tex] [0,\infty) \to [0-5,\infty-5) = [-5,\infty) [/tex]
Image of -3:
To compute [tex] f(-3) [/tex], simply plug [tex] x=-3 [/tex] in the formula:
[tex] f(-3) = 3(-3)^2-5 = 3\cdot 9 - 5 = 27-5 = 22 [/tex]
Numbers associated with 43:
We want to see which x value we must choose to get a y value of 43. So, the equation is
[tex] y = 3x^2-5 \to 43 = 3x^2-5 \iff 48 = 3x^2 \iff 16 = x^2 \iff x = \pm\sqrt{16} = \pm4 [/tex]
