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Quadratic functions are no exception when it comes to the types of transformations they can undergo. Applying a translation, reflection, stretch, or shrink to a quadratic function always leads to another quadratic function.
Rule

Rule

### Translation

By adding some number to every function value, $g(x) = f(x) + k,$ its graph is translated vertically. To instead translate it horizontally, a number is subtracted from the input of the function rule. $g(x) = f(x - h)$ The number $h$ is subtracted and not added, so that a positive $h$ translates the graph to the right.

Translate graph to the right

Translate graph upward

Notice that if the quadratic function $f(x) = ax^2$ is translated both vertically and horizontally, the resulting function is $g(x) = a(x - h)^2 + k.$ This is exactly the vertex form of a quadratic function. The vertex of $f(x) = ax^2$ is located at $(0,0).$ When the graph is then translated $h$ units horizontally and $k$ units vertically, the vertex moves to $(h, k).$

Rule

### Reflection

A function is reflected in the $x$-axis by changing the sign of all function values: $g(x) = \text{-} f(x).$ Graphically, all points on the graph move to the opposite side of the $x$-axis, while maintaining their distance to the $x$-axis.

Reflect graph in $x$-axis

A graph is instead reflected in the $y$-axis, moving all points on the graph to the opposite side of the $y$-axis, by changing the sign of the input of the function. $g(x) = f(\text{-} x)$ Note that the $y$-intercept is preserved.

Reflect graph in $y$-axis

Rule

### Stretch and Shrink

A function graph is vertically stretched or shrunk by multiplying the function rule by some constant $a > 0$: $g(x) = a \cdot f(x).$ All vertical distances from the graph to the $x$-axis are changed by the factor $a.$ Thus, preserving any $x$-intercepts.

Stretch graph vertically

By instead multiplying the input of a function rule by some constant $a > 0,$ $g(x) = f(a \cdot x),$ its graph will be horizontally stretched or shrunk by the factor $\frac 1 a.$ Since the $x$-value of $y$-intercepts is $0,$ they are not affected by this transformation.

Stretch graph horizontally

Exercise

Reflecting $f$ in the $x$-axis, and then adding $1$ to the input of the resulting function, gives $g.$ Determine which graph, I or II, corresponds to $g.$

Solution

By applying the transformations to the graph of $f,$ it will overlap with either I or II. This way, we can determine which graph is that of $g.$ The reflection in the $x$-axis moves every point on the graph to the other side of the $x$-axis.

Adding $1$ to the input of the resulting function gives us $g$: $g(x) = \text{-} f(x + 1).$ We can recognize this as a translation of $\,\text{-} \hspace{-1pt} f(x)$ by $1$ unit to the left. Graphing this translation, we see that the graph of $g$ coincides with graph II.

Thus, II is the graph of $g.$

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Exercise

The rules of $f$ and $g$ are given such that $g$ is a transformation of $f.$ $f(x) = 2x^2 \qquad g(x) = f(x - 1) + 2$ Describe the transformation(s) $f$ underwent to become $g.$ Then, write the rule of $g$ in vertex form and plot its graph.

Solution

Notice that subtracting $1$ from the input of $f$ and adding $2$ to the output gives the function $g.$ We can recognize the subtraction from the input as a translation to the right by $1$ unit. Adding $2$ to the output corresponds to a translation upward by $2$ units. Thus, $f$ has been translated $1$ unit to the right and $2$ units upward. The function $g$ is defined by $g(x) = f(x - 1) + 2,$ so we have to find $f(x - 1)$ to be able to state the rule of $g.$ This is done by replacing every $x$ in the rule of $f$ with $x - 1.$ $f(x) = 2x^2 \quad \Leftrightarrow \quad f(x - 1) = 2(x - 1)^2$ Substituting this into the rule of $g$ gives us $g(x) = 2(x - 1)^2 + 2.$ Notice that this function is already written in vertex form. To graph the function, we'll start by plotting the vertex, $(1,2),$ and the axis of symmetry.

Substituting $x = 0$ into the rule $g(x)$ gives us the $y$-intercept.

$g(x) = 2(x - 1)^2 + 2$
$g({\color{#0000FF}{0}}) = 2({\color{#0000FF}{0}} - 1)^2 + 2$
$g(0) = 2(\text{-} 1)^2 + 2$
$g(0) = 2 \cdot 1 + 2$
$g(0) = 2 + 2$
$g(0) = 4$

We can now plot the point $(0,4),$ and reflect it in the axis of symmetry at $(2, 4).$

Now, connecting the points with a parabola gives the desired graph.

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