What is the y-cooridinate for the point on the curve with x-cooridante 20?

Question

What is the y-coordinate for the point on the curve with x-coordinate 20?

$F. 160$

$G. 162$

$H. 164$

$J. 166$

$K. 168$

The explanation says "The correct answer is G. If the x-coordinate is 20, then the y-coordinate can be found by substituting 20 for x: $0.005\times20^2 - 200 = (0.005\times400) - 40 + 200 = 0.5(4) + 160 = 2+160$. In theory, you could read the value off the graph but you would not be able to read it accurately enough"

I don't understand how they got to this answer.. Is this a formula I don't know about? How would I calculate this answer on future problems?

I'm studying for the ACT so it's important I know how they got to this answer, not just the correct answer for one problem.

Answer 1

Each polynomial, which is a formula looking like $ax^n + bx^{n-1} +...$, is a map between the x-axis and the y-axis, where $y = f(x) = ax^n + bx^{n-1} + ... $, or, in your case, $f(x) = 0.005x^2 -2x - 200$. In order to calculate the value of $y$ at a given $x$, one applies the function defining this map to $x$, obtaining $f(x)$, which is $y$.

In order to apply a function $f(x) = ax^n + ...$, every instance of $x$ in the function body is substituted with the value of $x$ you are applying the function on, and the remaining expression is evaluated.

Edit: If you were not given the polynomial, or expected to derive it from a diagram or drawing of it, then there's no real way to arrive at the answer, besides guessing.

Answer 2

If there is a problem that states f(x), this means that there is some function of x, namely the one provided. This means that for every value of x (for example 1, 2, 3, PI, etc.) defined in the domain, there is an output. Graphically, this means you go to some point on the x-axis and follow it up or down until it meets the curve of f(x). f(x) is usually called y, which is the "height".

If there is no graph provided, but a function. And x is equal to something, replace every x with that thing. And it will tell you what f(x) is at that point.

In your case, y=f(x)=.005*(x)^2-2*(x)+200 and x=20. So, plugging in 20 wherever there is an x, y=f(20)=.005*(20)^2-2*(20)+200. This is equal to 162=y=f(20).