# f(x)=x^2-3It's for Functions.  How is this done?

pohnpei397 | Certified Educator

What this is saying is that f(x) is always going to be equal to x^2 - 3.  This will be true no matter what value x has.  In a mathematical function, one part of the function (called the value or the output) is totally determined by the other part of the function, which is called the argument or the input.  In this case, the value of f(x) is completely determined by the value of x.

So what you do here is you just plug numbers in for x.  If x = 0, for example, f(x) = -3. This is because 0^2 is 0 and 0-3 is -3.

If x = 2, f(x) = 1. This is because 2^2 = 4 and 4 - 3 = 1.

Depending on what you have been asked to do, you can make a table of the values for x and f(x) or you can make a graph.

hala718 | Certified Educator

If your question is How to solve this problem, and find out the X, f(x ) value and graph

f(x)= X^2-3

x^2= 3 ==> x= +- (3^(1/2))

and when x= 0 ==> f(0)= -3

to draw the graph you will need to draw a curve with the following points

(3^1/3, 0), (-3^1/2, 0), and  (0, -3)

help that helps

neela | Student

(Sice this to be a college level subject from your address, I would like to cover the answer analysing the perspective of the function and tracing it by the properties of the function).

f(x) = x^2-3 as a function represents parabola. It is a function of  second degree.

Also we can write y = x^2-3.

Symmetry: The curve is symmetrixcal about y axis as for x and -x  the function gives the same positive y value.

Vertex : The vertex of the cuve is  (0, -3)

The cuve is open upwards and goes for positive ifinite values on both left and right as x becomes large and large.

The curve, obviously ,has a value of -3 lowest when x=0. And that y= -3 ids the intercept value of the cuve on the Y axis.

The curve as crosses the x axis at equal distances of sqrt3 on the right and -sqrt3 on the left.

The curve y= x^2-3 Or x^2 = y+3or parabola has Could be compares to a

X^2 = 4aX a standard parabola. So we can write x^2 = 4(1/4) (y+y). So a= 1/4 is the focal length and (0, -3+1/4) or (0, -11/4 ) are the co ordinate positions of the focus. And the equation of the directrix is  y = -3-1/4 . Or y = -13/4 a parallel line to X axis .