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17

TRANSLATIONS

Translations of a parabola


A TRANSLATION OF A GRAPH is its rigid movement, vertically or horizontally.

    

On the left is the graph of the absolute value function.  On the right is its translation to a "new origin" at (3, 4).

The equation of the absolute value function is

y = |x|.

The equation of its translation to (3, 4) is

y − 4 = |x − 3|.

For, if we put x = 3,  then  y − 4 = 0, that is, y = 4.

In general,

If the graph of
 
  y  =  f(x)  
 
is translated a units horizontally and b units vertically,
 
then the equation of the translated graph is
 
  yb  =  f(xa).  

Example 1.   Write the equation of this graph:

 Answer.   y − 3 = |x + 5|.

The graph of the absolute value has been translated 3 units up, but 5 units to the left.  a = −5.  Therefore,  xa  becomes

x − (−5) = x + 5.

Problem 1.   Write the equation of this graph:

To see the answer, pass your mouse over the colored area.
To cover the answer again, click "Refresh" ("Reload").

y + 3 = −|x + 4|.

Not only has the graph of the absolute value been translated, it has also been reflected about the x-axis.

Problem 2.   Sketch the graph of  y = |x − 3|.

Problem 3.   Sketch the graph of  y = −|x + 2|.

Problem 4.   Sketch the graph of  y = −|x − 3| + 2.

This is equivalent to  y − 2 = −|x − 3|.

The graph is reflected about the x-axis, and translated to (3, 2).

Problem 5.   Sketch the graph of  y = .

This is the square root function translated 1 unit to the right.

Problem 6.   Sketch the graph of  y = −.

This is the reflected square root function, translated 3 units to the left.

Problem 7.   Sketch the graph of  y = 1 − x².

This is equivalent to  y − 1 = −x²,  which is the reflected parabola translated 1 unit up.

Example 2.  The vertex of a parabola.   Write the equation of the parabola (with leading coefficient 1) whose vertex is at the point (a, b).

 Answer.    yb = (xa)².   This is a translation of  y = x²  to (a, b).

Problem 8.   Write the equation of the parabola whose vertex is at

   a)  (1, 2)    y − 2 = (x − 1)²
 
   b)  (−1, 2)   y − 2 = (x + 1)²
 
   c)  (1, −2) y + 2 = (x − 1)²

Example 3.   What are the coördinates of the vertex of this parabola?

y  =  x² + 6x + 9
 
  Solution.   To answer, we must make the equation look like this --
 
yb  =  (xa

The vertex will then be at (a, b).

Now,  x² + 6x + 9  is the perfect square of  (x + 3):

y = x² + 6x + 9 = (x + 3)².

Therefore, a = −3, and b = 0.  The vertex is at (−3, 0.)

Example 4.   What are the coördinates of the vertex of this parabola?

y = x² + 5

 Solution.   Again, we must make the equation look like this:

yb = (xa)².

If we simply transpose 5 --

y − 5 = x²

-- we see that a = 0, and b = 5.  The vertex is at (0, 5).

Example 5.  Completing the square.   What are the coördinates of the vertex of this parabola?

y  =  x² + 6x −2
 
  Solution.   To make this form --
 
yb  =  (xa

-- we will transpose the constant term, and complete the square on the right.

y + 2  =  x² + 6x
 
  Complete the square by adding 9 to both sides:
 
y + 2 + 9  =  x² + 6x + 9
 
y + 11  =  (x + 3

The vertex is at (−3, −11).

Problem 9.   What are the coördinates of the vertex of this parabola?

y = x² − 10x + 25

The right-hand side is the perfect square of (x − 5).

y = (x − 5)²

The vertex therefore is at (5, 0).

Problem 10.   What are the coördinates of the vertex of this parabola?

y = x² − 1

The equation implies

y + 1 = x².

The vertex is at (0 −1).

Problem 11.   What are the coördinates of the vertex of this parabola?

y = x² − 8x + 1

Transpose the constant term, and complete the square on the right:

y − 1  =  x² − 8x
 
y − 1 + 16  =  x² − 8x + 16
 
y + 15  =  (x − 4)²

The vertex is at (4, −15).


The equation of a circle

What characterizes every point (x, y) on the circumference of a circle?

Every point is the same distance r from the center.  Therefore, according to the Pythagorean theorem,

x² + y² = r².

This is the equation of a circle of radius r, with center at the origin (0, 0).

Question.   What is the equation of a circle with center at (a, b) and radius r?

 Answer.   (xa)² + (yb)² = r².

The circle has been translated from (0, 0) to (a, b).

Problem 12.   Write the equation of the circle of radius 3, and center at the following point.

   a)   (1, 2)   (x − 1)² + (y − 2)² = 9
 
   b)   (−1, −2)   (x + 1)² + (y + 2)² = 9
 
   c)   (1, −2)   (x − 1)² + (y + 2)² = 9

Example 6.   Show that the following is the equation of a circle.  Name the radius and the coördinates of the center.

x² − 4x + y² − 2y = 11

 Solution.   To show that something is the equation of a circle, we must show that it can have this form,

(xa)² + (yb)² = r².

Therefore, we will complete the square in both x and y.

To complete the square in x, we will add 4 to both sides.

To complete the square in y, we will add 1 to both sides.

(x² − 4x + 4) + (y² − 2y + 1)  =  11 + 4 + 1
 
(x2)² + (y1  =  16.

This is the equation of a circle of radius 4, whose center is at (2, 1).

We can say then that when a quadratic in x plus a quadratic in y is equal to a number --

x² − 4x  +  y² − 2y = 11

-- then that is the equation of a circle.

The coefficients of x² and y² are 1. And the number must be greater than the negative of the sum of the squares of half the coefficients of x and y.

Problem 13.   Show that the following is the equation of a circle.  Name the radius and the coördinates of the center.

x² + 6x + y² + 10y − 2 = 0

Transpose the constant term, and complete the square in both x and y.  Add the same square numbers to both sides:

(x² + 6x + 9) + (y² + 10y + 25)  =  2 + 9 + 25
 
(x + 3)² + (y + 5)²  =  36

This is the equation of a circle of radius 6, with center at (−3, −5).

Vertical stretches and shrinks

If we multiply a function f(x) by a number c -- producing  c f(x) -- what will be the effect on the graph?

If we multiply f(x) by a number greater than 1 -- as in the graph in the center -- then every y-value is stretched.

But if we multiply f(x) by a number less than 1 -- as in the graph on the right -- then every y-value is shrunk.


Next Topic:  Rational functions


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