great READ-THROUGH through Dr. Carol JVF Burns (website creator) Follow together with the highlighted message while friend listen! Note: $\,f(x)+2\,$ is review aloud as: ‘$\,f\,$ the $\,x\,$ (slight pause) plus $\,2\,$’ $\,f(x+2)\,$ is review aloud as: ‘$\,f\,$ of (slight pause) $\,x\,$ plus $\,2\,$’ In this section, i over-emphasize the pauses, to aid you hear the difference.


You are watching: Graph that goes up and down

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$y = f(x)$ $y = f(x) + 2\,$ up $2$ $y = f(x)$ $y = f(x) - 2\,$ under $2$ $y = f(x)$ $y = f(x+2)\,$ left $2$ $y = f(x)$ $y = f(x-2)\,$ appropriate $2$
movements up and down adjust the $y$-values that points; transformations that impact the $y$-values room intuitive (e.g., to move up $\,2\,$, you add $\,2\,$ come the previous $\,y\,$-value) movements left and right adjust the $x$-values that points; changes that affect the $x$-values room counter-intuitive (e.g., to relocate left $\,2\,$, you change every $\,x\,$ by $\,x+2\,$, not $\,x-2\,$)
shifting up/down/left/right walk NOT change the shape of a graph.

The lesson Graphing Tools: Vertical and also Horizontal Translations in the Algebra II curriculum provides a thorough discussion of moving graphs up/down/left/right. The an essential concepts are recurring here. The practice in this great duplicate those in Graphing Tools: Vertical and Horizontal Translations.


points on the graph of $\,y=f(x)\,$ room of the form $\,\bigl(x,f(x)\bigr)\,$. point out on the graph that $\,y=f(x)+3\,$ space of the form $\,\bigl(x,f(x)+3\bigr)\,$. Thus, the graph of $\,y=f(x)+3\,$ is the very same as the graph of $\,y=f(x)\,$, shifted UP three units. Transformations entailing $\,y\,$ work the method you would suppose them to work—they space intuitive. below is the thought procedure you need to use once you are offered the graph the $\,y=f(x)\,$ and also asked around the graph the $\,y=f(x)+3\,$:
$$ \beginalign \textoriginal equation: &\quad y=f(x)\cr\cr \textnew equation: &\quad y=f(x) + 3 \endalign $$
$$ \begingather \textinterpretation of new equation:\cr\cr \overset\textthe new y-values\overbrace \strut\ \ y\ \ \overset\textare\overbrace \strut\ \ =\ \ \overset\quad\textthe ahead y-values\quad\overbrace \strut f(x) \overset\qquad\textwith 3 included to them\quad\overbrace \strut\ \ + 3\ \ \endgather $$
an overview of upright translations: allow $\,p\,$ it is in a hopeful number.
start with the equation $\,\colorpurpley=f(x)\,$. adding $\,\colorgreenp\,$ come the ahead $\,\colorgreeny\,$-values gives the new equation $\,\colorgreeny=f(x)+p\,$. This shifts the graph up $\,\colorgreenp\,$ units. A allude $\,\colorpurple(a,b)\,$ top top the graph that $\,\colorpurpley=f(x)\,$ moves to a point $\,\colorgreen(a,b+p)\,$ ~ above the graph the $\,\colorgreeny=f(x)+p\,$ .
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Additionally: start with the equation $\,\colorpurpley=f(x)\,$. individually $\,\colorgreenp\,$ native the ahead $\,\colorgreeny\,$-values gives the brand-new equation $\,\colorgreeny=f(x)-p\,$. This shifts the graph under $\,\colorgreenp\,$ units. A allude $\,\colorpurple(a,b)\,$ ~ above the graph of $\,\colorpurpley=f(x)\,$ moves to a suggest $\,\colorgreen(a,b-p)\,$ ~ above the graph the $\,\colorgreeny=f(x)-p\,$.
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This transformation form (shifting up and also down) is formally called vertical translation.
point out on the graph of $\,y=f(x)\,$ space of the type $\,\bigl(x,f(x)\bigr)\,$. point out on the graph that $\,y=f(x+3)\,$ are of the type $\,\bigl(x,f(x+3)\bigr)\,$. How can we find these preferred points $\,\bigl(x,f(x+3)\bigr)\,$? First, walk to the allude $\,\colorredP\bigl(x+3\,,\,f(x+3)\bigr)\,$ on the graph of $\,\colorredy=f(x)\,$. This suggest has the $\,y$-value that us want, however it has the wrong $\,x$-value. relocate this allude $\,\colorpurple3\,$ devices to the left. Thus, the $\,y$-value continues to be the same, yet the $\,x$-value is reduced by $\,3\,$. This gives the desired point $\,\colorgreen\bigl(x,f(x+3)\bigr)\,$. Thus, the graph of $\,y=f(x+3)\,$ is the very same as the graph the $\,y=f(x)\,$, shifted LEFT three units. Thus, instead of $\,x\,$ through $\,x+3\,$ relocated the graph LEFT (not right, as can have been expected!) Transformations involving $\,x\,$ do NOT occupational the means you would suppose them come work! They space counter-intuitive—they are against your intuition. right here is the thought procedure you have to use as soon as you are offered the graph that $\,y=f(x)\,$ and asked about the graph that $\,y=f(x+3)\,$:
$$ \beginalign \textoriginal equation: &\quad y=f(x)\cr\cr \textnew equation: &\quad y=f(x+3) \endalign $$
$$ \begingather \textinterpretation of brand-new equation:\cr\cr y = f( \overset\textreplace $x$ by $x+3$\overbrace x+3 ) \endgather $$
instead of every $\,x\,$ through $\,x+3\,$ in one equation moves the graph $\,3\,$ systems TO THE LEFT.
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an introduction of horizontal translations: permit $\,p\,$ it is in a confident number. start with the equation $\,\colorpurpley=f(x)\,$. change every $\,\colorgreenx\,$ by $\,\colorgreenx+p\,$ to provide the new equation $\,\colorgreeny=f(x+p)\,$. This move the graph LEFT $\,\colorgreenp\,$ units. A point $\,\colorpurple(a,b)\,$ top top the graph that $\,\colorpurpley=f(x)\,$ moves to a suggest $\,\colorgreen(a-p,b)\,$ ~ above the graph of $\,\colorgreeny=f(x+p)\,$.
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Additionally: begin with the equation $\,\colorpurpley=f(x)\,$. replace every $\,\colorgreenx\,$ by $\,\colorgreenx-p\,$ to provide the brand-new equation $\,\colorgreeny=f(x-p)\,$. This shifts the graph best $\,\colorgreenp\,$ units. A point $\,\colorpurple(a,b)\,$ ~ above the graph the $\,\colorpurpley=f(x)\,$ move to a suggest $\,\colorgreen(a+p,b)\,$ on the graph that $\,\colorgreeny=f(x-p)\,$.
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This transformation type (shifting left and also right) is formally called horizontal translation.

notice that different words are provided when talking about transformations entailing $\,y\,$, and also transformations involving $\,x\,$.

because that transformations including $\,y\,$ (that is, changes that adjust the $\,y$-values that the points), us say: carry out THIS to the vault $\,y$-value. for transformations involving $\,x\,$ (that is, transformations that change the $\,x$-values that the points), we say: replace the previous $\,x$-values by $\ldots$


upright translations: going indigenous $\,y=f(x)\,$ come $\,y = f(x) \pm c\,$

horizontal translations: going indigenous $\,y = f(x)\,$ come $\,y = f(x\pm c)\,$

Make sure you see the difference between (say) $\,y = f(x) + 3\,$ and also $\,y = f(x+3)\,$!

In the instance of $\,y = f(x) + 3\,$, the $\,3\,$ is ‘on the outside’; we"re dropping $\,x\,$ in the $\,f\,$ box, gaining the corresponding output, and then including $\,3\,$ to it. This is a vertical translation.

In the instance of $\,y = f(x + 3)\,$, the $\,3\,$ is ‘on the inside’; we"re including $\,3\,$ to $\,x\,$ before dropping it right into the $\,f\,$ box. This is a horizontal translation.


Question: begin with $\,y = f(x)\,$. relocate the graph come THE ideal $\,2\,$. What is the new equation?
Solution: This is a transformation involving $\,x\,$; it is counter-intuitive. You should replace every $\,x\,$ by $\,x-2\,$. The new equation is: $\,y = f(x-2)\,$
Solution: This is a transformation involving $\,y\,$; the is intuitive. You need to subtract $\,3\,$ indigenous the ahead $\,y\,$-value. The new equation is: $\,y = x^2 - 3\,$
Question: let $\,(a,b)\,$ it is in a allude on the graph the $\,y = f(x)\,$. Then, what point is ~ above the graph of $\,y = f(x+5)\,$?
Solution: This is a revolution involving $\,x\,$; it is counter-intuitive. replacing every $\,x\,$ by $\,x+5\,$ in an equation reasons the graph to transition $\,5\,$ devices to the LEFT. Thus, the new point is $\,(a-5,b)\,$.
grasp the concepts from this section by practicing the exercise at the bottom that this page. once you"re done practicing, move on to: horizontal and also vertical stretching/shrinking


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top top this exercise, you will certainly not key in your answer. However, you can inspect to view if your answer is correct. Note: there are lots of questions like this: “Start with $\,y = f(x)\,$. Relocate UP $\,2\,$. What is the brand-new equation?” right here is the very same question, stated an ext precisely: “Start with the graph that $\,y = f(x)\,$. Move this graph increase $\,2\,$.What is the equation the the brand-new graph?” every the ‘graphs’ are implicit in the trouble statements. (When miscellaneous is implicit climate it"s interpreted to it is in there, also though you can"t check out it.) problem TYPES: 1 2 3 4 5 6 7 8 9 10 11 12 13
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available MASTERED IN progression