Example 5.4.1.
First we complete this table
| \(x\) | \(x^2\) | |
| \(-2\) | \((-2)^2\) | \(=4\) |
| \(-1\) | \((-1)^2\) | \(=1\) |
| \(0\) | \(0^2\) | \(=0\) |
| \(1\) | \(1^2\) | \(=1\) |
| \(2\) | \(2^2\) | \(=4\) |
Next we graph these points and sketch the graph through them.
Section 5.4 Transformations of Quadratics
We have considered the nature of quadratic data and solving quadratic expressions. Now we will look at special properties of quadratics.
Subsection 5.4.1 Properties of Quadratics
First we will graph the most basic quadratic \(y=x^2\text{.}\) We will use this to compare all the variations that can be produced.
Notice that there is a single point at the bottom from which the parabola grows upward to the left and upward to the right. This is known as the vertex. It is at (0,0) for this parabola.
Notice as well that the left and right sides are mirrors of each other. Specifically they are mirrored over the line through the vertex known as the line of symmetry. In this case that is the vertical line \(x=0\text{.}\)
Example 5.4.2.
We will graph \(y=2(x-1)^2-4\text{.}\) First we complete this table
| \(x\) | \(2(x-1)^2-4\) | |
| \(-2\) | \(2(-2-1)^2-4\) | \(=14\) |
| \(-1\) | \(2(-1-1)^2-4\) | \(=4\) |
| \(0\) | \(2(0-1)^2-4\) | \(=-2\) |
| \(1\) | \(2(1-1)^2-4\) | \(=-4\) |
| \(2\) | \(2(2-1)^2-4\) | \(=-2\) |
Next we graph these points and sketch the graph through them.
Notice that this time the vertex is at \((1,-4)\text{.}\) The line of symmetry is \(x=1\text{.}\)
Subsection 5.4.2 Transformations
Checkpoint 5.4.3.
In this exercise we will check on the effect of multiplying a quadratic by a scalar (number).
(a)
Complete the table below. Compare the results to the table in Example 5.4.1
| \(x\) | \(2x^2\) |
| \(-2\) | \(2(-2)^2=8\) |
| \(-1\) | |
| \(0\) | |
| \(1\) | |
| \(2\) |
(b)
Graph these points. Compare the graph to the graph in Example 5.4.1
Checkpoint 5.4.4.
In this exercise we will continue to check on the effect of multiplying a quadratic by a scalar (number).
(a)
Complete the table below. Compare the results to the table in Example 5.4.1
| \(x\) | \(\frac{1}{2}x^2\) |
| \(-2\) | \(\frac{1}{2}(-2)^2=2\) |
| \(-1\) | |
| \(0\) | |
| \(1\) | |
| \(2\) |
(b)
Graph these points. Compare the graph to the graph in Example 5.4.1
Checkpoint 5.4.5.
In this exercise we will check on the effect of adding inside the square.
(a)
Complete the table below. Compare the results to the table in Example 5.4.1
| \(x\) | \((x-1)^2\) |
| \(-2\) | \((-2-1)^2=9\) |
| \(-1\) | |
| \(0\) | |
| \(1\) | |
| \(2\) |
(b)
Graph these points. Compare the graph to the graph in Example 5.4.1
Checkpoint 5.4.6.
In this exercise we will continue to check on the effect of adding inside the square.
(a)
Complete the table below. Compare the results to the table in Example 5.4.1
| \(x\) | \((x+1)^2\) |
| \(-2\) | \((-2+1)^2=1\) |
| \(-1\) | |
| \(0\) | |
| \(1\) | |
| \(2\) |
(b)
Graph these points. Compare the graph to the graph in Example 5.4.1
Checkpoint 5.4.7.
In this exercise we will check on the effect of adding outside the square.
(a)
Complete the table below. Compare the results to the table in Example 5.4.1
| \(x\) | \(x^2-1\) |
| \(-2\) | \((-2)^2-1=3\) |
| \(-1\) | |
| \(0\) | |
| \(1\) | |
| \(2\) |
(b)
Graph these points. Compare the graph to the graph in Example 5.4.1
Checkpoint 5.4.8.
In this exercise we will continue to check on the effect of adding outside the square.
(a)
Complete the table below. Compare the results to the table in Example 5.4.1
| \(x\) | \(x^2+1\) |
| \(-2\) | \((-2)^2+1=5\) |
| \(-1\) | |
| \(0\) | |
| \(1\) | |
| \(2\) |
(b)
Graph these points. Compare the graph to the graph in Example 5.4.1
Checkpoint 5.4.9.
In this exercise we will check on the effect of multiplying a quadratic by a negative.
(a)
Complete the table below. Compare the results to the table in Example 5.4.1
| \(x\) | \(-x^2\) |
| \(-2\) | \(-(-2)^2=-4\) |
| \(-1\) | |
| \(0\) | |
| \(1\) | |
| \(2\) |
(b)
Graph these points. Compare the graph to the graph in Example 5.4.1
Subsection 5.4.3 Forms
We have solved quadratics using the (standard) form \(ax^2+bx+c=0\) and graphed parabolas using the form \(a(x-h)^2+k\text{.}\) From the form we used for graphing we can easily identify the vertex. With some effort it is possible to find the vertex from the standard form. We will look at a method that does not require using more algebra than we have practiced here.
Example 5.4.10.
Find the vertex of \(y=2x^2-11x+12\text{.}\)
Solution.
Because of the symmetry of a parabola we know that the vertex lies half way between any pair of points with the same height. For example it is half way between the solutions to \(2x^2-11x+12 = 0\text{.}\)
\begin{align*}
x & = \frac{-(-11) \pm \sqrt{(-11)^2-4(2)(12)}}{2(2)}\\
& = \frac{11 \pm \sqrt{121-96}}{4}\\
& = \frac{11 \pm \sqrt{25}}{4}\\
& = \frac{11 \pm 5}{4}\\
& = 4, 3/2.
\end{align*}
To find what is half way in between we average these two values.
\begin{align*}
x & = \frac{4+3/2}{2}\\
& = \frac{8/2+3/2}{2}\\
& = 11/4
\end{align*}
Thus the x-coordinate of the vertex is 11/4. To find the y-coordinate we substitute the x value into the quadratic.
\begin{equation*}
2(11/4)^2-11(11/4)+12 = -25/8.
\end{equation*}
