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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.

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 $(h,k).$

A function is reflected in the $x$-axis by changing the sign of all function values: $g(x)=-f(x).$ 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. $g(x)=f(-x)$ Note that the $y$-intercept is preserved.

Reflect graph in $y$-axis

A function graph is vertically stretched or shrunk by multiplying the function rule by some constant $a>0$: $g(x)=a⋅f(x).$ 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 $a>0,$ $g(x)=f(a⋅x),$ its graph will be horizontally stretched or shrunk by the factor $a1 .$ Since the $x$-value of $y$-intercepts is $0,$ they are not affected by this transformation.

Stretch graph horizontally

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 $g.$

Show 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 $g$: $g(x)=-f(x+1).$ We can recognize this as a translation of $-f(x)$ 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 $g.$

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

Show 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 $x−1.$
$f(x)=2x_{2}⇔f(x−1)=2(x−1)_{2}$
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.

$g(x)=2(x−1)_{2}+2$

Substitute$x=0$

$g(0)=2(0−1)_{2}+2$

SubTermSubtract term

$g(0)=2(-1)_{2}+2$

CalcPowCalculate power

$g(0)=2⋅1+2$

MultiplyMultiply

$g(0)=2+2$

AddTermsAdd terms

$g(0)=4$

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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