Graphing Functions

Graphing Functions exercise

• Describe how the given graph represents a function.

  Hints

  We 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 y-intercept.

  Each function is a relation but not each relation is a function.

  Solution

  Functions 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 y-intercept $8$; 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.

  Hints

  Functions 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.

  Solution

  To 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 right-most 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 $y=2x^2+3$.

  Hints

  For 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.

  Solution

  Here 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.

  Hints

  You 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 $1$.

  Solution

  To 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:

  $\begin{array}{c|r|r|r|r|r|r|r|r|r} x &-10 & -9 & -8 & -7 & -6 & - 5& -4 & -3 & -2 & -1 & 0 \\ \hline y & 10 & 8.1 & 6.4 & 4.9 & 3.6 & 2.5 & 1.6 & 0.9 & 0.4 & 0.1 & 0 \end{array}$

  $\begin{array}{c|r|r|r|r|r|r|r|r|r} x & 0 & ~~~1 &~~~ 2 & ~~~3 &~~~4 & ~~~5& ~~~6 & ~~~7 & ~~~8 & ~~~9 & ~~~10 \\ \hline y & 0 & 0.1 & 0.4 & 0.9 & 1.6 & 2.5 & 3.6 & 4.9 & 6.4 & 8.1 & 10 \end{array}$

  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 $y=2x+8$.

  Hints

  Here 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 $y=2\times 3+8=6+8=14$.

  Solution

  To 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.

  Hints

  A 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 y-intercept.

  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

  $y=a(x-v_x)^2+v_y$.

  Solution

  Here 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 y-intercept $5$ 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 $y=x^2-4$.
  • 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 $y=x^2+1$.