Graphing Functions
Description Graphing Functions
Graphing functions is pretty much the same as graphing equations, with the symbol f(x) being a fancy way of writing y. To graph a function f(x), follow the following steps:
(1) Pick some values for x. (2) Substitute in each value of x into f(x); the result will be the y value of the x which was plugged in. (3) Plot the resulting ordered pairs (x,y) in the plane. (4) If the equation is linear, connect the points with a line, otherwise connect the points with a smooth curve.
The domain (i.e. the x values) gives the vertical projection of the graph, whereas the range (i.e. the y values) gives the horizontal projection of the graph.
To determine if the graph is a function or not, we use the vertical line test: if any vertical line touches only one point of the graph, then the graph represents a function; this means graphically that every x is assigned to exactly one y. Otherwise, the graph does not represent a function.
Analyze functions using different representations. CCSS.MATH.CONTENT.HSF.IF.C.7
Transcript Graphing Functions
Stephanie Gawking is a math enthusiast, and her hobby is astronomy. From her backyard, she gazes through her telescope and dreams of discovering a new celestial body.
She sees a shooting star  which is a small, fast meteor. Although she knows the average speed of a shooting star is 30,000 miles per hour, she wonders about its path and how far it'll travel in a given amount of time.
What is a function?
To show the relationship between distance and time, we can look at a graph; the path of the star MAY be the graph of a function, but how do we know for sure?
A function is a special relationship between two variables; in this case, the variables are the distance and the time. For each minute that passes, the star travels to a new location in the sky. If the graph of the star's path is a function, then for every input, time, there is a unique output, the location or the distance traveled. Let’s take a look at the function f(x) = 2x + 8. Notice that we used the function notation, f(x), this is just a fancy way of representing 'y'.
Graphing a function
Okay, let’s graph the function. It’s already written in slopeintercept form, y = mx + b. The yintercept is equal to 8, and the slope, or rise over run, is equal to 2.
OR, you can write the values for 'x' and 'y' in a table. For instance, when x = 0, y = 8. When x = 1, y = 10, and so on. Next, plot a few points and connect the points on the line. We know the graph displays a function because each 'x' has only one 'y'. To double check that the graph is a function, we can also do a vertical line test. Draw in several vertical lines, if the lines touch the graph in only one place, then the graph is a function. If any line touches the graph in more than one place, it's not a function  it's that simple!
Graphing Parabolas
Let’s graph y = x^{2}. To do this, we can create a function table and calculate a few points, then graph. If x = 2, then y = 4. If x = 1, y = 1, and so on. Notice the shape of this function. This distinctive ushape is called a parabola. When you have a quadratic equation, the graph is always a parabola. How do we know if a quadratic equation is a function? For each x, there is only one y, and the graph passes the vertical line test.
Vertical Line Test
Looking through her telescope, Stephanie sees a constellation. It’s so curvy; is it a function? Let’s use the vertical line test to see if it is. Oh look! It passes the test! So this graph is also a function  for each input, 'x', there is one output, 'y'. Here’s another awesome constellation, but does its graph form a function? Because the graph passes the vertical line test, it sure does!
And, what about this one? It’s ushaped, but it’s turned sideways. It fails the vertical line test, so no, it’s not a function. For each input, there is more than one output. Whoa nelly! This one looks like a circle. Is it a function? It fails the test, so no way.
Stephanie adjusts her telescope. Holy moly! Stop the presses! What's that? She thinks she's finally discovered a new celestial body. It’s a dream come true. Wait, is that a firefly?
Graphing Functions Exercise

Describe how the given graph represents a function.
HintsWe have that $f(x)=2x+8$ is a function, as well as $f(x)=8x+2$.
A linear function in slope intercept form is given by
$f(x)=mx+b$,
where $m$ is the slope while $b$ is the yintercept.
Each function is a relation but not each relation is a function.
SolutionFunctions are very special relations between variables. In particular, for us, they are relations between two variables, say $x$ and $y$, where every $x$ is related to at most one $y$.
For example, $f(x)=2x+8$ is a linear function with the slope $2$ and the yintercept $8$; the corresponding graph is a line.
Given just a graph, how can we check if it represents a function? Well, we can use the vertical line test! Which says that a graph represents a function if any vertical line has only one point in common with the graph.
By drawing vertical lines over the graph of $f(x)=2x+8$, we can see that it passes the vertical line test and thus represents a function!

Decide which graphs represent functions.
HintsFunctions are special relations between variables. In particular, for us, they are relations between two variables, say $x$ and $y$, where every $x$ is related to at most one $y$.
Use the vertical line test: a graph represents a function if any vertical line has only one point in common with the graph.
SolutionTo check if a graph represents a function, you can use the vertical line test. The vertical line test says that a graph represents a function if any vertical line has only one point in common with the graph.
For example $x^2+y^2=16$ is a circle with radius $4$. This equation is not a function, as its graph does not pass the vertical line test.
We can see that the graph of the sideways parabola $x=y^2$ (i.e. the bottom rightmost picture) also does not pass the vertical line test and thus does not represent a function.
All the other graphs represent functions, as they pass the vertical line test.

Complete the function table for the function $y=2x^2+3$.
HintsFor each $x$ you can calculate the corresponding $y$ by plugging $x$ into the function.
When $x=4$, we have that $y=2(4)^2+3=2\times 16+3=32+3=35$.
Pay attention to the sign of $x$. A negative number squared is a positive number.
SolutionHere you see the resulting graph as well as the three points $(0,0)$, $(1,5)$, and $(1,5)$.
$~$
Any point lying on the graph is given by $(x,y)$, where $y=2x^2+3$.
$~$
We can construct a function table by taking some $x$values and calculating their corresponding $y$values:
 For $x=3$ we get $y=2(3)^2+3=2\times 9+3=18+3=21$
 For $x=2$ we get $y=2(2)^2+3=2\times 4+3=8+3=11$
 For $x=1$ we get $y=2(1)^2+3=2\times 1+3=2+3=5$
 For $x=0$ we get $y=2(0)^2+3=2\times 0+3=0+3=3$
 For $x=1$ we get $y=2(1)^2+3=2\times 1+3=2+3=5$
 For $x=2$ we get $y=2(2)^2+3=2\times 4+3=8+3=11$
 For $x=3$ we get $y=2(3)^2+3=2\times 9+3=18+3=21$

Decide which stars belong to the graph of the function.
HintsYou can check each star by plugging its $x$coordinate into the function. The resulting $y$value must match the $y$coordinate of that star for that star to be part of the constellation.
Plugging $x=0$ into the function, we get $y=\frac{1}{10}(0)^2=0$.
So we can see that the star at $(0,0)$ belongs to the graph.
The star at $(4,1)$ does not belong to the graph since
$y=f(4)=\frac{1}{10}(4)^2=1.6$, not $1$.
SolutionTo check if a star lies in our constellation, we need to plug in its $x$coordinate into the function $f(x)=\frac{1}{10} x^2$ and see if the resulting $y$coordinate is indeed the $y$coordinate of that star. We can construct a function table to help figure out which stars are in the constellation:
$\begin{array}{crrrrrrrrr} x &10 & 9 & 8 & 7 & 6 &  5& 4 & 3 & 2 & 1 & 0 \\ \hline y & 10 & 8.1 & 6.4 & 4.9 & 3.6 & 2.5 & 1.6 & 0.9 & 0.4 & 0.1 & 0 \end{array}$
$\begin{array}{crrrrrrrrr} x & 0 & ~~~1 &~~~ 2 & ~~~3 &~~~4 & ~~~5& ~~~6 & ~~~7 & ~~~8 & ~~~9 & ~~~10 \\ \hline y & 0 & 0.1 & 0.4 & 0.9 & 1.6 & 2.5 & 3.6 & 4.9 & 6.4 & 8.1 & 10 \end{array}$
Graphing the points in this table, we see that the stars which belong to the constellation are: $(0,0),(10,10),(10,10),(4,1.6),(6,3.6)$, and $(5,2.5)$.

Explain how to draw the graph of the function $y=2x+8$.
HintsHere you see how to plot the point $(3,6)$ in a coordinate system.
You can write a linear function as $f(x)=2x+8$, as well as $y=2x+8$.
The $y$ corresponding to $x=3$ is $y=2\times 3+8=6+8=14$.
SolutionTo draw the graph of a function, first we need to draw a coordinate system with a horizontal line, the $x$axis, and a vertical one, the $y$axis.
You can either draw a line corresponding to a linear function by drawing the $y$intercept and then use the slope to determine the other points on the line (Remember: the slope is given by rise over run), or you can figure out a function table of $(x,y)$ pairs.
To get these pairs, plug each $x$ value into the function; for example, the $y$ coordinate at $x=0$ is $f(0)=2\times 0+8=8$. So the pair we get is $(0,8)$.
From the function table, the graph of the function can be drawn by plotting the points from the function table on the coordinate system and drawing a line connecting all of the points.

Match the graph with its corresponding function.
HintsA line is the graph of a linear function.
A linear function in slope intercept form is given by
$y=mx+b$,
where $m$ is the slope and $b$ the yintercept.
The graph of a quadratic function is a parabola.
If you know the vertex $(v_x, v_y)$ of a parabola you can write the corresponding equation as
$y=a(xv_x)^2+v_y$.
SolutionHere you have two lines and two parabolas.
We first want to remember that:
 A line corresponds to a linear function.
 A parabola corresponds to a quadratic function.
Let's start with the lines:
 The line on the left has the yintercept $5$ as well as this one on the right. But what is the difference? It's the slope!
 The line on the lieft has a negative slope while the line on the right has a positive slope.
 You get the slope using "rise over run": for the left line we have a rise of $5$ and a run of $2$. So the corresponding function is $y=\frac52x+5$.
 The rise of the right line is $5$ and the run is $2$. This gives us the function $y=\frac52x+5$.
Now for the parabolas:
 The left parabola has intercepts the $y$axis at $(0,4)$.
 So the left parabola is the parabola $y=x^2$ shifted down by $4$, and the corresponding function is $y=x^24$.
 The right parabola has intercepts the $y$axis at $(0,1)$.
 So the right parabola is the parabola $y=x^2$ shifted up by $1$, and the corresponding function is $y=x^2+1$.