the most basic quadratic function:

$y=X^2$

the most basic quadratic function:

$y=X^2$

The family of parabolas $y=x^2+c$

The basic quadratic function – with the addition of $c$

In this family, we are given a quadratic function that clearly shows us how the function moves horizontally – how many steps it needs to move right or left.

$P$ represents the number of steps the function will move horizontally – right or left.

If $P$ is positive – (there is a minus in the equation) – the function will move $P$ steps to the right.

If $P$ is negative – (and as a result, there is a plus in the equation because minus times minus equals plus) – the function will move $P$ steps to the left.

In this quadratic function, we can see a combination of horizontal and vertical shifts:

$K$: Determines the number of steps and the direction the function will move vertically – up or down.

$K$ positive – shift up, $K$ negative – shift down.

$P$: Determines the number of steps and the direction the function will move horizontally – right or left.

What is the value of y for the function?

\( y=x^2 \)

of the point \( x=2 \)?

Meet the most basic quadratic function:

$y=X^2$

In this function:

$b=0~, c=0~, a=1$

The function is a smiling minimum function and its vertex is - $(0,0)$

The axis of symmetry of this function is $X=0$.

The increasing interval of the function – all the $X$ values where the function is increasing are: $X>0$

The decreasing interval of the function – all the $X$ values where the function is decreasing are: $X<0$

Positive interval: all $X$ except $0$ – you can see in the graph that the entire function is above the $X$ axis

Negative interval: none. The entire function is above the $X$ axis.

Let's continue to a similar function from the same family:

$y=-x^2$

In this function:

$a=-1 ,~b=0 ,~c=0$

The function is a sad maximum function and its vertex is $(0,0)$

The axis of symmetry of this function is $X=0$.

Increasing interval: $X<0$

Decreasing interval: $X>0$

Positive interval: None. The entire function is below the $X$ axis.

Negative interval: All $X$ except $X=0$

Let's continue with another function from the same family:

$y=ax^2$

In this function:

$a=some~number,~b=0,~ c=0$

Its vertex is- $(0,0)$

The axis of symmetry of this function is $X=0$.

The larger $a$ is: the parabola will have a smaller opening – closer to its axis of symmetry.

The smaller $a$ is: the parabola will have a larger opening – farther from its axis of symmetry.

The function from this family has no horizontal or vertical shift because in every function from this family the vertex is $(0,0)$

Click here to practice and learn more about the function $y=X^2$

The family of parabolas $y=x^2+c$

The basic quadratic function – with the addition of $c$

In this function:

$c$ – represents the intersection point of the function with the $Y$ axis.

The meaning of $c$ is a vertical shift up or down of the function from the vertex $(0,0)$.

If $c$ is positive – the function will shift vertically up by the number of steps indicated by $c$.

If $c$ is negative – the function will shift vertically down by the number of steps indicated by $c$.

This function describes only vertical shifts up and down according to $c$

Let's see an example:

$y=4x^2+7$

In this function -

$C=7$

this means that the intersection point of the function with the $Y$ axis is $7$.

And actually – the function moves vertically $7$ steps up.

Click here to practice and learn more about the function $y=X^2+c$

In this family, we are given a quadratic function that clearly shows us how the function moves horizontally – how many steps it needs to move right or left.

$P$ represents the number of steps the function will move horizontally – right or left.

If $P$ is positive – (there is a minus in the equation) – the function will move $P$ steps to the right.

If $P$ is negative – (and as a result, there is a plus in the equation because minus times minus equals plus) – the function will move $P$ steps to the left.

Let's see an example:

$Y=(X-6)^2$

This function moves from the vertex $(0,0)$,$6$ steps to the right

Let's see this in the figure:

Click here to practice and learn more about the function $y=(x-p)^2$

Test your knowledge

Question 1

Find the ascending area of the function

\( f(x)=2x^2 \)

Question 2

One function

\( y=6x^2 \)

to the corresponding graph:

Question 3

One function

\( y=-6x^2 \)

to the corresponding graph:

In this quadratic function, we can see a combination of horizontal and vertical shifts:

$K$: Determines the number of steps and the direction the function will move vertically – up or down.

$K$ positive – shift up, $K$ negative – shift down.

$P$: Determines the number of steps and the direction the function will move horizontally – right or left.

Let's look at an example of combining two translations together:

For example, in the function:

$y=(x+2)^2+5$

The changes will be:

According to $P=-2$: the parabola will move $2$ steps to the left.

According to $K=5$: the parabola will move $5$ steps up.

Let's see this in the illustration:

We can see that the vertex of the parabola is:

$(-2,5)$

Click here to practice and learn more about the function $y=(x-p)^2$

Do you know what the answer is?

Question 1

One function

\( y=-2x^2-3 \)

to the corresponding graph:

Question 2

Which chart represents the function \( y=x^2-9 \)?

Question 3

Find the descending area of the function

\( f(x)=\frac{1}{2}x^2 \)

What is the positive domain of the function below?

$y=(x-2)^2$

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, $x\ne2$

What is the value of y for the function?

$y=x^2$

of the point $x=2$?

$y=4$

Find the ascending area of the function

$f(x)=2x^2$

0 < x

One function

$y=6x^2$

to the corresponding graph:

2

One function

$y=-6x^2$

to the corresponding graph:

4

Related Subjects

- Quadratice Equations and Systems of Quadraric Equations
- Quadratic Equations System - Algebraic and Graphical Solution
- Solution of a system of equations - one of them is linear and the other quadratic
- Intersection between two parabolas
- Word Problems
- Properties of the roots of quadratic equations - Vieta's formulas
- Ways to represent a quadratic function
- Various Forms of the Quadratic Function
- Standard Form of the Quadratic Function
- Vertex form of the quadratic equation
- Factored form of the quadratic function
- The quadratic function
- Quadratic Inequality
- Parabola
- Symmetry in a parabola
- Plotting the Quadratic Function Using Parameters a, b and c
- Finding the Zeros of a Parabola
- Methods for Solving a Quadratic Function
- Completing the square in a quadratic equation
- Squared Trinomial
- The quadratic equation
- Families of Parabolas
- The functions y=x²
- Family of Parabolas y=x²+c: Vertical Shift
- Family of Parabolas y=(x-p)²
- Family of Parabolas y=(x-p)²+k (combination of horizontal and vertical shifts)