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Transformations

In this lesson, we will determine the different types of Transformations that can be applied to the parent function \(y = x^2\) by expressing it in the Vertex Form, \(y = a(x-h)^2 + k\).

Reflection, Stretching and Compression

The \(a\) value will both determine whether the quadratic function will get either stretched or compressed and which direction it will open:

  • If \(|a| > 1\), the graph will get stretched by a factor of \(a\). This will make the graph appear narrower, as the \(x\) value will have a greater effect on the \(y\) value
  • If \(|a| < 1\), the graph will get compressed by a factor of \(a\). This will make the graph appear wider, as the \(x\) value won't have as great of an effect on the \(y\) value
  • If \(a > 0\), the function will be positive, making the parabola open upward
  • If \(a < 0\), the function will be negative, reflecting the graph and making the parabola open downward

Quadratic stretched by a factor of 2, making it appear more narrow.
Quadratic compressed by a factor of -1/3, making it appear wider and flipping it on its x-axis.

State the transformation(s) in the quadratic equation \(y = -6x^2\)

We can first identify \(a = -6\). This means:

  • The graph is vertically stretched by a factor of \(6\)
  • The graph is reflected in the \(x\)-axis

Horizontal Shifts

The \(h\) value will determine which direction horizontally the graph will get shifted. This change can be represented algebraically by the equation \(y = (x-h)^2\):

  • If \(h > 0\), the graph will get shifted to the right by \(h\) units
Quadratic shifted 2 units to the right.

  • If \(h < 0\), the graph will get shifted to the left by \(h\) units
Quadratic shifted 4 units to the left.

Identify the quadratic equation for a graph that is shifted \(5\) units left

Since the graph is getting shifted to the left, we can identify \(h = -5\)

When we plug the value into the equation, the negatives cancel each other out. As a result, we get the final equation:

\(y = (x + 5)^2\)

Vertical Shifts

The \(k\) value will determine which direction vertically the graph will get shifted. This change can be represented algebraically by the equation \(y = x^2 + k\):

  • If \(k > 0\), the graph will get shifted upwards \(k\) units
Quadratic shifted 5 units upward.
  • If \(k < 0\), the graph will get shifted downwards \(k\) units
Quadratic shifted 4 units downward.

Identify the quadratic equation for a graph that is shifted \(7\) units upward

Since the graph is getting shifted to the upward, we can identify \(k = 7\).

When we plug the \(k\)-value into the equation, we get the final equation:

\(y = x^2 + 5\)
Identify and graph a qudratic equation that is reflected, compressed by a factor of \(1/4\), shifted \(3\) units right and shifted \(2\) units downward

First, we can identify how the following transformations can be applied to the parent function:

  • Since the graph is reflected and stretched, we can identify \(a = -1/4\)
  • Since the graph is getting shifted to the right, we can identify \(h = 3\)
  • Since the graph is getting shifted downward, we can identify \(k = -2\)

When we plug all the values into the equation, we get the final equation:

\(y = -\cfrac{1}{4}(x-3)^2 - 2\)

In order to assist with graphing, we can create a table of values (from \(x=-1\) to \(x =7\)) to help determine where each of the points lie:

x Values -1 0 1 2 3 4 5 6 7
y Values -6 -4.25 -3 -2.25 -2 -2.25 -3 -4.25 -6


We can now sketch our graph as such:

Graph of a parabola representing the quadratic equation y=-1/4(x-3)²-2.
Identify the formula of the following graph.
Parabola shifted 4 units to the left and 3 units upward, compressed by a factor of 0.5.

We can represent the formula of this quadratic in vertex form:

\(y = a(x-h)^2 + k\)

First, we can identify the vertex of the quadratic as \((-4, 3)\). We can plug this point into the equation:

\(y = a(x+4)^2 + 3\)

Next, we can pick another point the quadratic intersects with. In this case, we will choose \((-6, 5)\). We can substitute this point into the formula to solve for \(a\):

\(5 = a(-6+4)^2 + 3\)

\(5-3 = a(-2)^2\)

\(2 = 4a\)

\(a = 0.5\)

Now that we have determined \(a = 0.5\), we can write our formula as such:

\(y = 0.5(x+4)^2 + 3\)

Therefore, we can represent the algebraically represent the function as \(\boldsymbol{y = 0.5(x+4)^2 + 3}\)



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