great READ-THROUGH by Dr. Carol JVF Burns (website creator) Follow in addition to 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 read aloud as: ‘$,f,$ that (slight pause) $,x,$ plus $,2,$’ In this section, ns over-emphasize the pauses, to aid you listen the difference.


You are watching: How to shift a graph down

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$y = f(x)$ $y = f(x) + 2,$ up $2$ $y = f(x)$ $y = f(x) - 2,$ down $2$ $y = f(x)$ $y = f(x+2),$ left $2$ $y = f(x)$ $y = f(x-2),$ right $2$
activities up and down change the $y$-values of points; changes that affect the $y$-values space intuitive (e.g., to relocate up $,2,$, you include $,2,$ come the vault $,y,$-value) activities left and also right adjust the $x$-values the points; revolutions that affect the $x$-values are counter-intuitive (e.g., to relocate left $,2,$, you change every $,x,$ by $,x+2,$, not $,x-2,$)
moving up/down/left/right go NOT readjust the shape of a graph.

The lesson Graphing Tools: Vertical and also Horizontal Translations in the Algebra II curriculum gives a thorough discussion of changing graphs up/down/left/right. The crucial concepts are recurring here. The exercises in this class duplicate those in Graphing Tools: Vertical and Horizontal Translations.


point out on the graph that $,y=f(x),$ room of the type $,igl(x,f(x)igr),$. point out on the graph the $,y=f(x)+3,$ are of the type $,igl(x,f(x)+3igr),$. Thus, the graph of $,y=f(x)+3,$ is the exact same as the graph that $,y=f(x),$, shifted UP 3 units. Transformations involving $,y,$ occupational the means you would suppose them to work—they space intuitive. below is the thought process you have to use as soon as you are given the graph the $,y=f(x),$ and asked around the graph the $,y=f(x)+3,$:
$$ eginalign extoriginal equation: &quad y=f(x)crcr extnew equation: &quad y=f(x) + 3 endalign $$
$$ egingather extinterpretation of brand-new equation:crcr overset extthe new y-valuesoverbrace strut y overset extareoverbrace strut = oversetquad extthe previous y-valuesquadoverbrace strut f(x) oversetqquad extwith 3 included to themquadoverbrace strut + 3 endgather $$
an overview of vertical translations: let $,p,$ be a positive number.
begin with the equation $,colorpurpley=f(x),$. adding $,colorgreenp,$ to the ahead $,colorgreeny,$-values provides the brand-new equation $,colorgreeny=f(x)+p,$. This move the graph increase $,colorgreenp,$ units. A allude $,colorpurple(a,b),$ top top the graph of $,colorpurpley=f(x),$ moves to a allude $,colorgreen(a,b+p),$ top top the graph that $,colorgreeny=f(x)+p,$ .
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Additionally: begin with the equation $,colorpurpley=f(x),$. individually $,colorgreenp,$ indigenous the previous $,colorgreeny,$-values provides the brand-new equation $,colorgreeny=f(x)-p,$. This shifts the graph down $,colorgreenp,$ units. A point $,colorpurple(a,b),$ ~ above the graph of $,colorpurpley=f(x),$ moves to a suggest $,colorgreen(a,b-p),$ top top the graph the $,colorgreeny=f(x)-p,$.
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This transformation form (shifting up and down) is formally dubbed vertical translation.
points on the graph that $,y=f(x),$ are of the kind $,igl(x,f(x)igr),$. point out on the graph of $,y=f(x+3),$ room of the type $,igl(x,f(x+3)igr),$. How can we locate these wanted points $,igl(x,f(x+3)igr),$? First, go to the point $,colorredPigl(x+3,,,f(x+3)igr),$ top top the graph that $,colorredy=f(x),$. This point has the $,y$-value that we want, but it has the wrong $,x$-value. move this allude $,colorpurple3,$ units to the left. Thus, the $,y$-value continues to be the same, however the $,x$-value is decreased by $,3,$. This offers the desired allude $,colorgreenigl(x,f(x+3)igr),$. Thus, the graph of $,y=f(x+3),$ is the same as the graph of $,y=f(x),$, shifted LEFT 3 units. Thus, replacing $,x,$ by $,x+3,$ relocated the graph LEFT (not right, as could have to be expected!) Transformations involving $,x,$ do NOT occupational the method you would intend them come work! They room counter-intuitive—they are against your intuition. here is the thought procedure you have to use when you are provided the graph of $,y=f(x),$ and asked about the graph that $,y=f(x+3),$:
$$ eginalign extoriginal equation: &quad y=f(x)crcr extnew equation: &quad y=f(x+3) endalign $$
$$ egingather extinterpretation of brand-new equation:crcr y = f( overset extreplace $x$ by $x+3$overbrace x+3 ) endgather $$
instead of every $,x,$ through $,x+3,$ in an equation moves the graph $,3,$ devices TO THE LEFT.
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an overview of horizontal translations: let $,p,$ it is in a optimistic number. begin with the equation $,colorpurpley=f(x),$. change every $,colorgreenx,$ by $,colorgreenx+p,$ to give the new equation $,colorgreeny=f(x+p),$. This move the graph LEFT $,colorgreenp,$ units. A allude $,colorpurple(a,b),$ top top the graph that $,colorpurpley=f(x),$ move to a point $,colorgreen(a-p,b),$ top top the graph that $,colorgreeny=f(x+p),$.
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Additionally: begin with the equation $,colorpurpley=f(x),$. change every $,colorgreenx,$ by $,colorgreenx-p,$ to give the new equation $,colorgreeny=f(x-p),$. This shifts the graph appropriate $,colorgreenp,$ units. A suggest $,colorpurple(a,b),$ ~ above the graph the $,colorpurpley=f(x),$ moves to a suggest $,colorgreen(a+p,b),$ ~ above the graph that $,colorgreeny=f(x-p),$.
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This transformation kind (shifting left and right) is formally dubbed horizontal translation.

notice that different words are provided when talking around transformations involving $,y,$, and also transformations including $,x,$.

because that transformations entailing $,y,$ (that is, revolutions that adjust the $,y$-values that the points), us say: perform THIS to the previous $,y$-value. for transformations involving $,x,$ (that is, revolutions that readjust the $,x$-values that the points), we say: change the previous $,x$-values by $ldots$


upright translations: going from $,y=f(x),$ to $,y = f(x) pm c,$

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

Make certain you view the difference between (say) $,y = f(x) + 3,$ and also $,y = f(x+3),$!

In the situation of $,y = f(x) + 3,$, the $,3,$ is ‘on the outside’; we"re dropping $,x,$ in the $,f,$ box, acquiring the equivalent output, and then adding $,3,$ to it. This is a upright translation.

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


Question: start with $,y = f(x),$. relocate the graph to THE ideal $,2,$. What is the brand-new equation?
Solution: This is a change involving $,x,$; the is counter-intuitive. You must 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 should subtract $,3,$ native the previous $,y,$-value. The new equation is: $,y = x^2 - 3,$
Question: let $,(a,b),$ it is in a allude on the graph that $,y = f(x),$. Then, what allude is on the graph of $,y = f(x+5),$?
Solution: This is a transformation involving $,x,$; that is counter-intuitive. instead of every $,x,$ through $,x+5,$ in one equation causes the graph to transition $,5,$ units to the LEFT. Thus, the new point is $,(a-5,b),$.
grasp the concepts from this ar by practicing the practice at the bottom of this page. as soon as you"re done practicing, relocate on to: horizontal and also vertical stretching/shrinking


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top top this exercise, you will not key in your answer. However, you can check to watch if your answer is correct. Note: over there are numerous questions like this: “Start through $,y = f(x),$. Move UP $,2,$. What is the new equation?” right here is the very same question, stated an ext precisely: “Start v the graph of $,y = f(x),$. Relocate this graph up $,2,$.What is the equation that the new graph?” all the ‘graphs’ space implicit in the difficulty statements. (When miscellaneous is implicit then it"s taken to be there, also though girlfriend can"t view it.) difficulty TYPES: 1 2 3 4 5 6 7 8 9 10 11 12 13
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obtainable MASTERED IN progression