# Family of Parabolas y=(x-p)²

🏆Practice parabola of the form y=(x-p)²

### Family of Parabolas $y=(x-p)^2$

In this family, we have a slightly different quadratic function that shows us, very clearly, how the parabola shifts horizontally.
$P$ indicates the number of steps the parabola will move horizontally, to the right or to the left.
If $P$ is positive: (there is a minus sign in the equation) - The parabola will move $P$ steps to the right.
If $P$ is negative: (and, consequently, there will be a plus sign in the equation since minus by minus equals plus) - The parabola will move $P$ steps to the left.

Let's see an example:
The function  $Y=(X+2)^2$

shifts two steps to the left.
Let's see it in an illustration:

## Test yourself on parabola of the form y=(x-p)²!

Find the intersection of the function

$$y=(x-2)^2$$

With the X

## Change in Slope

When there is an $a$ before the parentheses, it indicates the slope of the parabola.
The larger the $a$, the closer the parabola will be to its axis of symmetry. Steeper - with a smaller opening.
The smaller the $a$, the farther the parabola will be from its axis of symmetry. Less steep - with a larger opening.

Let's see an example that shows the balance between the change in slope and horizontal shift:
For example, in the function

$Y=\frac{1}{2} (X-2)^2$

the changes will be:
Shift of $2$ steps to the right
and the change in slope, The parabola will be less steep and with a larger opening
Let's see it in an illustration:

### Graphical and Algebraic Solution when$y=0$

Algebraic Solution

when $y=0$ the expression inside parentheses must be equal to $0$.
$X$ must be equal to $P$ for the equation to be correct.

## Graphical Solution

The graphical solution is the vertex of the parabola.
In a quadratic function of this form
$Y=(X-p)^2$
in which the coefficient of $X^2$  is $1$,
the vertex of the parabola is composed of$Y=0$  and by$P$ which indicates the $X$ vertex.
$(0,P)$
Note that, when in the equation there is a minus sign before the$P$
indeed, it is positive and, when there is a plus sign before the $P$ it is, in fact, negative.

## Examples and exercises with solutions from the family of parabolas y=(x-p)²

### Exercise #1

What is the positive domain of the function below?

$y=(x-2)^2$

### Step-by-Step Solution

In the first step, we place 0 in place of Y:

0 = (x-2)²

We perform a square root:

0=x-2

x=2

And thus we reveal the point

(2, 0)

This is the vertex of the parabola.

Then we decompose the equation into standard form:

y=(x-2)²

y=x²-4x+2

Since the coefficient of x² is positive, we learn that the parabola is a minimum parabola (smiling).

If we plot the parabola, it seems that it is actually positive except for its vertex.

Therefore the domain of positivity is all X, except X≠2.

all x, $x\ne2$

### Exercise #2

Find the intersection of the function

$y=(x-2)^2$

With the X

### Video Solution

$(2,0)$

### Exercise #3

Find the intersection of the function

$y=(x+4)^2$

With the Y

### Video Solution

$(0,16)$

### Exercise #4

To work out the points of intersection with the X axis, you must substitute $x=0$.

False

### Exercise #5

To find the y axis intercept, you substitute $x=0$ into the equation and solve for y.