The graph of a function
We know that the graph of a function is the set of all points of the plan xOy that have like abscissa the faces (elements of the domain sets) and like ordinate they have the corresponding value of the function.
The graph of the numeric function is the illustration of the graph on the coordinative plan.
The graph of the numeric function f:$\displaystyle X\to R$ in the coordinative plan xOy, is the set of all points (x,f(x)) where x∈X.
If we have the graph of a numeric function on the coordinative plan xOy we can find:
1. The domain set X. This is the set of all the abscissa points.
2. For the element a∈X we can find the corresponding value f(a) of the function.
3. The set of the values of the function.
Example 1: We have two lines:
a) Which of them represents a graph of a numeric function? Why?
We should draw vertical lines to see if any of them intersects our graph in more then one point.We see that on the first graph we find a vertical line that intersects the graph on two points,so our first graph is not a numeric function.Meanwhile in the second graph we can´t find a line that intersects our graph in more then one point,so our second graph is a numeric function.Each vertical line we can draw intersects the graph on one and only one point.
b) For this numeric function find from the graph the domain and the range.
The horizontal line on the graph is from 1 to 5 so the domain of f is [1,5]
The vertical line on the graph is from -2 to 3 so the range of f is [-2,3]
c) Find from the graph ,f(-2) and f(3).
x =-2 is not included on the [1,5] domain of our function.
f(3) = -2 and from the graph we see that is the vertex of parabola C(3,-2).
Example 2: We have the numeric function f: y=x3, x∈[-2,2]. Are A(-1,1), B(2,8) and C(3,27) points of our graph?
Solution:
A(-1,1) $\displaystyle \to $ Replace the point at the function $ \displaystyle 1={{(-1)}^{3}}$.
$ \displaystyle 1\ne -1$, so the point A is not a point of our graph.
B(2,8) $\displaystyle \to (8)={{(2)}^{3}}$.8 = 8, so the point B is a point on our graph since it satisfies the equation of the graph
C(3,27) $\displaystyle \to $ this is a point on the graph of the function y=x3 x∈R but since x∈[-2,2] then this is not a point of our graph.
The graph of some known function
You know the graph of the linear function y = ax + b, x∈R is a line not parallel with the y-axis. To build this we only need two points of the line.The graph of the function y=ax+b, x∈X (where x is a subset of R) is a set of points that we can find on this line.
The graph of inverse proportion $ \displaystyle y=\frac{a}{x}$, x∈ (a≠0) is a curved line (hiperbola) made from 2 parts.
When a>0, one of this parts is located on the first quadrant and the other on the third quadrant.When a<0 the parts are located one on the second quadrant and the other on the fourth quadrant.
The graph of the function $ \displaystyle y=a{{x}^{2}}$, x∈R ( where a≠0) is a curved line(parabola) that has as a axis of symmetry the axis Oy and like vertex point the origin O.
When a>0 this parabola is located on the upper part of the plan and her arms go up in infinity. When a<0 this parabola is located on the lower part of the plan and her arms go down in infinity.
Example 3:
a) Create the graph of the function: y = -2x + 4, x ∈ R.
b) Which is the graph of the function y = -2x + 4, x ∈ [-1,3]?
Solution
a) 1. Draw a table to find points (x,y).
2. Draw the points on the coordinative plan and join them with a straight line.
b) 1. For x ∈ [-1,3], we find the points x,y.
2. Since the domain is x ∈ [-1,3] then the graph of the function is as below.
How to create the graph of the function y = ax2 + bx + c, x∈R.
We know that this function y = ax2 + bx + c, x ∈ R represents a curve called parabola. Were the vertex is the point C(m,n) where $ \displaystyle ~m=~-\frac{b}{{2a}}$ and $ \displaystyle n=\frac{D}{4}$
So to create the graph we follow the steps:
1. Find the axis of symmetry.
2. Find the vertex.
3. Determine if the arms will be up or down.
4. On a tabular form find points (x,y) by taking values up and down the value of axis.
5. Draw the points (x,y) on the quadratic plan.
6. Join the points with a curved line.
1. We first determine the coefficents a = 1. b = 6, c = 5 , then we use the formula $ \displaystyle ~m=~x=-\frac{b}{{2a}}$
$\displaystyle ~~x=-\frac{b}{{2a}}=-\frac{6}{2}=-3$
x = -3
2. C(m,n) or C(x,y) $\displaystyle \to C(\frac{{-b}}{{2a}},\frac{{-D}}{{4a}})$
D=b2-4ac=36-20=16
$ \displaystyle \frac{-D}{4a}=-\frac{16}{4}=-4$
C(-3, -4)
3. If a>0 then the arms of parabola will be up and if a<0 then the arms will be down.
In our case a=1, a>0 so the arms will be up.
4. Find points (x,y) by taking values up and down the value of axis, x=-3.
5 & 6.
