# How is the shift negative positive or zero

## Displacement of the normal parabola in the direction of the x-axis

This page is about shifting the normal parabola to the right or left. ### Equation of the shifted normal parabola

A parabolic equation of the form \$ f (x) = (x-d) ^ 2 \$ usually causes more problems in the clear interpretation than the equation \$ f (x) = x ^ 2 + c \$. Therefore, let's take a look at a table of values ​​using a specific example:

\$ \ begin {array} {l | c | c | c | c | c | c | c | c | c} x & -3 & -2 & -1 & 0 & 1 & 2 & 3 & 4 & 5 \ \ hline f_1 (x) = x ^ 2 & \ color {# 13f} {9} & \ color {# f00} {4} & \ color {# 1a1} {1} & \ color {# f61} {0} & \ color {# 18f} {1} & \ color {# b1f} {4} & \ color {# a61} {9} & 16 & 25 \ \ hline f_2 (x) = (x-2) ^ 2 & 25 & 16 & \ color {# 13f} {9} & \ color {# f00} {4 } & \ color {# 1a1} {1} & \ color {# f61} {0} & \ color {# 18f} {1} & \ color {# b1f} {4} & \ color {# a61} {9 } \ end {array} \$

Compared to the output function, with \$ f_2 (x) = (x-2) ^ 2 \$ all values ​​are two units behind right shifted, not to the left, which one might initially suspect because of the negative sign in the case of the two. For this reason, a minus is also used in the initial formula \$ f (x) = (x-d) ^ 2 \$ in order to be able to use the parameter \$ d \$ with the "correct" sign.

And this is how it looks (use slider to change):

The following applies to the graph of the quadratic function \$ f (x) = (x-d) ^ 2 \$:
The normal parabola is shifted by \$ d \$ in the direction of the \$ x \$ axis, namely to the right for positive \$ d \$ and to the left for \$ d <0 \$. The vertex \$ S (x_s | y_s) \$ has the coordinates \$ S (d | 0) \$, that is, \$ x_s = d \$ and \$ y_s = 0 \$.

Perhaps the best way to remember the opposite sign in the function equation is to focus on the vertex:
In the initial parabola with the equation \$ f (x) = x ^ 2 \$, the vertex lies in the coordinate origin \$ S (0 | 0) \$. If you move the parabola in the direction of the \$ x \$ axis, the \$ y \$ coordinate of the vertex does not change, i.e. it remains zero. We only achieve this for \$ x = d \$, because then \$ f (d) = (d-d) ^ 2 = 0 ^ 2 = 0 \$.

### Point test

As with straight lines, one also checks here whether a point lies on a parabola by inserting the coordinates into the associated functional equation.

example 1: Is the point \$ P (\ color {# f00} {- 3} | \ color {# 1a1} {4}) \$ on the graph of \$ f (x) = (x-1) ^ 2 \$?

solution: There are two possible solutions:

• You insert both coordinates and check whether a true statement is made:
\$ \ begin {align *} (\ color {# f00} {- 3} -1) ^ 2 & = \ color {# 1a1} {4} \ (-4) ^ 2 & = 4 \ 16 & = 4 && \ text {wrong statement} \ end {align *} \$
Since a wrong statement was made, the point does not lie on the parabola.
• You just insert the \$ x \$ coordinate and then compare with the given \$ y \$ coordinate:
\$ f (\ color {# f00} {- 3}) = (\ color {# f00} {- 3} -1) ^ 2 = (- 4) ^ 2 = 16 \ not = \ color {# 1a1} { y_p} \ Rightarrow P \$ does not lie on the parabola.

If a true statement had been made or if the function value had agreed with \$ y_p \$, then the point would be on the parabola.

Example 2: How must \$ x \$ be chosen so that the point \$ P (\ color {# f00} {x} | \ color {# 1a1} {9}) \$ on the graph of the function \$ f (x) = (x + 2) ^ 2 \$ lies?

solution: We insert the given quantities and solve for \$ x \$. As a solution, I have chosen immediate root extraction. You can of course break the brackets and use the \$ pq \$ formula.

\$ \ begin {align *} (\ color {# f00} {x} +2) ^ 2 & = \ color {# 1a1} {9} && | \ sqrt {{} \ phantom {6}} \ x + 2 & = 3 && \ text {or} & x + 2 & = - 3 && | -2 \ x_1 & = 1 &&& x_2 & = - 5 \ end {align *} \$

The points \$ P_1 (1 | 9) \$ and \$ P_2 (-5 | 9) \$ meet the condition.

### Determine parabolic equation

As with the parabola shifted in the \$ y \$ direction, there are two ways to define the equation. In my experience, the second type of task rarely (not?) Occurs in school, but I will present an example for interested students.

Example 3: The normal parabola is shifted three units to the left. Give your equation.

solution: Since the parabola should be shifted to the left, \$ d \$ is negative, so \$ d = \ color {# f00} {- 3} \$. So the equation is
\$ f (x) = (x - (\ color {# f00} {- 3})) ^ 2 \ f (x) = (x + 3) ^ 2 \$

Example 4: A normal parabola shifted in the direction of the \$ x \$ axis goes through the point \$ P (\ color {# f00} {5} | \ color {# 1a1} {4}) \$. Find a possible equation.

The type of task is extremely rare. Let's look at the task:

For this position of the point \$ P \$ there are two possible parabolas that fulfill the condition.

solution: We choose the approach \$ f (x) = (x-d) ^ 2 \$. With the point sample we can determine the possible values ​​for the parameter \$ d \$:

\$ \ begin {align *} (\ color {# f00} {5} -d) ^ 2 & = \ color {# 1a1} {4} && | \ sqrt {\ phantom {{} 6}} \ 5-d & = \ pm 2 \ 5-d & = 2 && \ text {or} & 5-d & = - 2 && | -5 \ -d & = - 3 &&& - d & = - 7 && |: (- 1) \ d_1 & = 3 &&& d_2 & = 7 \ f_1 (x) & = (x-3) ^ 2 &&& f_2 (x) & = (x-7) ^ 2 \ end {align *} \$

Here, too, I chose immediate root extraction as the solution technique because it is faster with the given shape. If this method does not suit you, you can of course also work with the \$ pq \$ formula.

Exercises

Last update: 02.12.2015; © Ina de Brabandt

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