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Name the 4 different types of shifts that are possible with a quadratic equation.
How is it expressed in the equation?





if a quadratic equation is graphed with a shift upward, how is that expressed in its equation? It is shown as x^2 + x, or x^2 + 1, or x^2 – x, or x^2 – 1.
The shifts of a quadratic equation are expressed in each equation through a formula. Please provide those formulas with a corresponding brief description of what does that equation represent. For example, if you provide a formula for a quadratic equation that shifts upward, then give the formula and then state it is a quadratic equation that shifts upward.

Another example, if you have
• f(x) = x2 + 1 upward 6 units.

Then its answer is (x^2 +1) + 6 simplified to x^2 + 7.
Or

If you have
• f(x) = (x + 6)2 to the right 10 units.


How is this shown in an equation? (if you know the formula, you will know how to do it that way as well?

Comments

f(x) = X^2
 
 f(x) - b = X^2
 
   f(x) = X^2 +b   is  shift of f(x) b units upward.
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2 Answers

f(x ) = ax^2 ± bX + c
 
        = a ( X^2 ± b/ a + c/a)
 
          = a ( X^2± 2 b/2a  + b^2/4 a^2 - b^2 /4a^2 + c/a)=
 
            a ( X ± b/2a) ^2   = b^2/ 4a^2 - c/a
 
             a ( X ± b/2a) ^2 = ( b^2 - 4ac)/ 4a^2     (1)
 
                Consider Y = X^2, or - X^2
 
                 Horizontal shift of  ±b/2a , and vertical shift of ( b^2-4ac / 4a^2) equation ( 1) will be  
                obtained.
 
               line   X = - b/2a   is a axis of symmetry.
 
                  The coordinate of Vertex ( -b/2a, ( b^2 - 4ac)/ 4a^2)
 
                   The Roots X -intercepts :
 
                    - b/ 2a + √(b^2 -4ac) / 2a , +b/2a  - √(b^2 -4ac) / 2a
 
                       are Symmetric about X = -b/2a
 
                      The roots are algebraically obtained from solution of equation (1) , the quadratic formula.
 
                      Take an example of
 
                   f(x)  =  3 X^2 + 5X - 9                             
 
                             = 3 ( X^2/ 3 + 5/3X - 3)     
                             
                              = 3 ( X^2 + 2( 5/6 ) X + 25/36 - 25/36 -3)
 
                                = ( X + 5/6 ) ^2 -  133/ 36
 
                                Vertex ( -5/6 , 133/36) 
 
                                 X = -5/6   Axis of Symmetry
                               
                               Roots:
 
                             ( X + 5/6 ) ^2  -133/36 =0
 
                                 X1 = -5/6 + √133/ 6      X2= -5/6 - √133/6
 
 
                  
 
 
 
 
 
 
 
                 
Name the 4 different types of shifts that are possible with a quadratic equation. How are they expressed in the equation?

1. Shift right a units: x → x – a
2. Shift left b units: x → x – (-b) = x + b
3. Shift up c units: y → y – c
4. Shift down d units: y → y – (-d) = y + d

where “→” means “changed to”. And these rules are true for ANY equation, not just a quadratic function; e.g.,
<h,k>: x^2 + y^2 = r^2
→ (x–h)^2 + (y–k)^2 = r^2.
 
<h,k> is a “translation vector” that says "shift h units in x-direction and k units in y-direction".

If a quadratic equation is graphed with a shift upward, how is that expressed in its equation?

It is shown as
x^2 + x,
x^2 + 1, ⇐ this is x^2 shifted up 1
x^2 – x,
or x^2 – 1.

The shifts of a quadratic equation are expressed in each equation through a formula. Please provide those formulas with a corresponding brief description of what does that equation represent. For example, if you provide a formula for a quadratic equation that shifts upward, then give the formula and then state it is a quadratic equation that shifts upward.

Another example, if you have
• f(x) = x^2 + 1 upward 6 units.

Then its answer is (x^2 +1) + 6 simplified to x^2 + 7.
Or

If you have
• f(x) = (x + 6)^2 to the right 10 units.

How is this shown in an equation? (if you know the formula, you will know how to do it that way as well)

f(x) = (x + 6)^2 to the right 10 units.

<10,0>: f(x) → g(x) = f(x-10)
g(x) = ((x–10) + 6)^2 = (x – 4)^2

f(x) = (x + 6)^2 to the left 10 units.

<–10,0>: f(x) → g(x) = f(x-(–10)) = f(x+10)
g(x) = ((x+10) + 6)^2 = (x – 4)^2

f(x) = (x + 6)^2 up 10 units.

<0,10>: f(x) → g(x) – 10 = f(x),
g(x) = f(x) + 10 = (x + 6)^2 + 10

f(x) = (x + 6)^2 down 10 units.

<0,–10>: f(x) → g(x) – (–10) = f(x),
g(x) = f(x) – 10 = (x + 6)^2 – 10
 
f(x) = (x + 6)^2 left 3 and down 5.
 
<-3,-5>: f(x) → g(x) – (-5) = f(x – (-3)),
g(x) = f(x – (-3)) + (-5) = f(x + 3) – 5 = ((x + 3) + 6)^2 – 5
g(x) = (x + 9)^2 – 5