Graphing Quadratic Functions

Graphing Quadratic Functions exercise

• Describe how the coefficients change the graph.

  Hints

  Let $b=c=0$ and let $a$ vary to see what happens. For example:

  $f(x)=x^2$ or $f(x)=3x^2$ ...

  Keep in mind:

  • $a$ controls the width of the parabola and if it opens downwards or upwards.
  • $b$ causes a shift along the $x$-axis.
  • $c$ causes a shift along the $y$-axis.

  The vertex is the lowest or the highest point of a parabola, depending on if the parabola is opened downwards or upwards.

  Solution

  Let's consider the parabolas in the picture to the right:

  The impact of the coefficient $a$

  Let's start with the green parabola, which belongs to $f(x)=x^2$; so $a=1$ and $b=c=0$. This parabola is upwards opened as $a>0$. If $a<0$ the parabola is downwards opened: for example, the blue one belonging to $f(x)=-x^2$.

  The coefficient $a$ impacts the graph in another way: the greater the absolute value of $a$ the thinner the parabola. Let's have a look at the yellow one, $f(x)=-\frac12x^2$. This parabola is wider than the blue one, or in other words, the blue one is thinner than the yellow one.

  The impact of the coefficient $b$

  The coefficient $b$ causes a shift of the vertex. Let's have a look at the violet parabola belonging to $f(x)=-x^2-4x-3$. Because $a=-1$ and $b=-4$ have the same sign, the vertex is shifted to the left. For $a=1$ we can conclude: for $b>0$ ($b<0$) the vertex is shifted to the left (right).

  The impact of the coefficient $c$

  The red parabola belongs to $f(x)=x^2+2$ and it is shifted upwards. So for all $c>0$ the parabola is shifted upwards and for $c<0$ downwards.

• Explain the change of the axis of symmetry depending on $a$ and $b$.

  Hints

  The lowest or the highest point of a parabola is called the vertex.

  The line parallel to the $y$-axis, passing through the vertex, is the axis of symmetry of the parabola.

  The $x$-coordinate of the vertex is given by $x=-\frac b{2a}$.

  Determine the sign of this $x$-coordinate depending on the signs of $a$ as well as of $b$.

  Solution

  The lowest, for upwards opened parabolas, or the highest, for downwards opened parabolas, point is called the vertex. The line parallel to the $y$-axis, passing this vertex is the axis of symmetry of the corresponding parabola.

  The $x$-coordinate of the vertex is given by $x=-\frac b{2a}$.

  Thus we can conclude that for $b=0$ the vertex lies on the $y$-axis and thus this axis is also the axis of symmetry.

  Let us consider the following cases:

  1. $a>0$ and $b>0$. This gives us $x<0$, a negative $x$-coordinate of the vertex $\rightarrow$ the axis of symmetry is shifted to the left.
  2. $a>0$ and $b<0$ leads to a positive $x$-coordinate and thus to a right-shift.
  3. $a<0$ and $b>0$ leads to a positive $x$-coordinate as well and also to a right-shift.
  4. $a<0$ and $b<0$ gives us a negative $x$-coordinate, that means a shift to the left of the axis of symmetry.

• Determine the corresponding function.

  Hints

  If the vertex is lying on the $y$-axis, we have $b=0$.

  If $a>0$ ($a<0$) the parabola is opened upwards (downwards).

  Changing $c$ causes a shift upwards or downwards depending on the sign of $c$.

  You can also check the vertex each time.

  Solution

  How can we determine the corresponding function for each parabola?

  First look at how it opens: is it opened upwards or downwards? If the parabola is opened upwards we have $a>0$ and $a<0$ otherwise. So the green, red, and orange parabolas have a positive $a$ and the blue one has a negative $a$.

  Next we look at the vertex:

  • If the $x$-coordinate is zero, we can conclude that $b=0$. So the blue, the green, and the red parabolas have $b=0$, while for the orange one $b\neq 0$.
  • If the vertex is $(0,0)$ the corresponding equation is given by $f(x)=ax^2$, otherwise it is given by $ax^2+c$, where $c$ is the $y$-intercept.

  Last we decide the width of the parabola, checking some points:

  • Let's start with the blue one: $a<0$ and $b=c=0$. Because $(1,-1)$ is one point on the parabola, we can conclude the corresponding function is $f(x)-x^2$.
  • The green one has $a>0$ and $b=c=0$. Let's check one point $(1,2)$, so $x=1$ and $y=2$. This is fulfilled for $f(x)=2x^2$.
  • The yellow one is the only parabola with a vertex not lying on the $y$-axis. Thus $b\neq 0$. Let us check the point $(1,0)$: $f(1)=1^2-1=0$, as desired. So $f(x)=x^2-x$ is the corresponding function.
  • Last but not least, the red one with the vertex $(0,3)$; so $a>1$, $b=0$m and $c=3$. The only function fulfilling this is $f(x)=x^2+3$.

• Decide which function belongs to which graph.

  Hints

  Note that you don't have to pay attention to the width of the parabola.

  For $a>0$ ($a<0$) the parabola is opened upwards (downwards).

  Pay attention to the vertex: if the vertex lies on the $y$-axis, you have that $b=0$.

  You can also check points lying on the parabola: if you put the $x$-coordinate in the function you get the $y$-coordinate.

  Solution

  Let's start with the grey and blue parabolas, which have a vertex on the $y$-axis; the corresponding equation is $f(x)=ax^2+c$.

  For each of these parabolas, $a$ is either $1$ and opened upwards, or $-1$ and opened downwards.

  The regarding $c$ is the $y$-intercept. So, for the grey parabola we have $f(x)=-x^2+1$ and for the blue we have $f(x)=x^2+2$.

  Next we investigate the green and violet parabolas, with the vertex lying on the $x$-axis. Both parabolas have $a=1$. Because the $x$-coordinate of the vertex is $-2$ for the green parabola and $2$ for the violet one, we can determine $b$ as follows:

  • For the green one, $-2=-\frac{b}{2a}$. Keeping in mind that $a=1$, we get $b=4$. $c$ is the $y$-intercept, namely $4$. Thus we get the resulting function $f(x)=x^2+4x+4$.
  • For the violet one, $2=-\frac{b}{2a}$. Keeping in mind that $a=1$, we get $b=-4$. $c$ is the $y$-intercept, namely $4$. Thus we get the resulting function $f(x)=x^2-4x+4$.

  There are still two parabolas to consider:

  • The red one has the vertex $(-4,2)$ and is upwards opened. Using this vertex we can conclude $f(x)=(x+4)^2+2=x^2+8x+16+2=x^2+8x+18$.
  • The yellow one has the vertex $(3,3)$ and is downwards opened. Using this vertex we can conclude $f(x)=-(x-3)^2+3=-x^2+6x-9+3=-x^2+6x-9$.

  For all the parabolas you can also check some points lying on the parabola. For example, for the yellow one $(2,2)$ we have

  $f(2)=-2^2+6(2)-6=-4+12-6=2$ $~~~~~$✓

• Check the following statements.

  Hints

  Here you see a parabola where the vertex has a positive $x$-coordinate and a negative $y$-coordinate. The green line is the axis of symmetry.

  Here you see a few parabolas.

  This is a linear function, which has a line as a graph.

  Solution

  A quadratic function in standard form is given by

  $f(x)=ax^2+bx+c$.

  Let $a=1$ and $b=c=0$.

  We then get $f(x)=x^2$, a special quadratic function.

  The graph of a quadratic function is a parabola. Beside you see a picture of different parabolas.

  A parabola can be upwards, or downwards, opened depending on the sign of $a$:

  • upwards for $a>0$
  • downwards for $a<0$

  The highest or the lowest point of the parabola is called the vertex. This could be the origin $(0,0)$, like for the green, yellow, and blue parabolas. This could also be any other point. For example, $(-2,1)$ for the violet parabola.

  Each parabola has a line parallel to the $y$-axis, passing the vertex, which is its axis of symmetry.

• Describe how to determine the vertex.

  Hints

  Determine the $x$-coordinate, $x=-\frac b{2a}$, first.

  If you know the $x$-coordinate, you get the $y$-coordinate by pugging the $x$-coordinate into the function.

  Let's have a look at an example: $f(x)=2x^2-2x+1$. Here we have $b=-2$.

  1. $x=-\frac{-2}{2(2)}=\frac12$
  2. $y=2\left(\frac12\right)^2-2\left(\frac12\right)+1=\frac12-1+1=\frac12$.

  So we can determine the vertex:

  $\left(\frac12,\frac12\right)$

  Solution

  The quadratic function we are considering is $f(x)=2x^2+bx+1$.

  What is the impact of changing values for $b$ on the vertex?

  The vertex is the highest or lowest point of the parabola.

  The $x$-coordinate is given as $x=-\frac{b}{2a}=-\frac b4$.

  We start with $b=2$:

  1. $x=-\frac24=-\frac12$.
  2. Now we put this $x$ in $f(x)=2x^2+2x+1$ and get $y=2\left(-\frac12\right)^2+2\left(-\frac12\right)+1=\frac12-1+1=\frac12$.

  The vertex is given by $\left(-\frac12,\frac12\right)$.

  In a similar manner we can determine the vertex for $b=4$:

  1. $x=-\frac44=-1$.
  2. Now we put this $x$ into the function $f(x)=2x^2+4x+1$ and get $y=2\left(-1\right)^2+4\left(-1\right)+1=2-4+1=-1$.

  So, the vertex is given by $\left(-1,-1\right)$.

  Lastly we figure out the vertex for $b=-4$:

  1. $x=-\frac{-4}4=1$.
  2. Now we put this $x$ in the functional equation $f(x)=2x^2-4x+1$ and get $y=2\left(1\right)^2-4\left(1\right)+1=2-4+1=-1$.

  So, the vertex is given by $\left(1,-1\right)$.

  Now the twins have all the information they need to catch the animals they want for their zoo.