the most basic quadratic function:

$y=X^2$

Question Types:

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.

Question 1

What is the positive domain of the function below?

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

Question 2

What is the value of y for the function?

\( y=x^2 \)

of the point \( x=2 \)?

Question 3

Find the ascending area of the function

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

Question 4

One function

\( y=6x^2 \)

to the corresponding graph:

Question 5

One function

\( y=-6x^2 \)

to the corresponding graph:

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

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

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

Question 4

Find the intersection of the function

\( y=(x+4)^2 \)

With the Y

Question 5

Find the intersection of the function

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

With the X

One function

$y=-2x^2-3$

to the corresponding graph:

4

Which chart represents the function $y=x^2-9$?

4

Find the descending area of the function

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

x < 0

Find the intersection of the function

$y=(x+4)^2$

With the Y

$(0,16)$

Find the intersection of the function

$y=(x-2)^2$

With the X

$(2,0)$

Question 1

Complete:

The missing value of the function point:

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

\( f(?)=16 \)

Question 2

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

Question 3

To find the y axis intercept, you substitute **\( x=0 \)** into the equation and solve for y.

Question 4

To which chart does the function \( y=x^2 \) correspond?

Question 5

One function

\( y=-x^2 \)

for the corresponding chart

Complete:

The missing value of the function point:

$f(x)=x^2$

$f(?)=16$

$f(4)$$f(-4)$

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

False

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

True

To which chart does the function $y=x^2$ correspond?

2

One function

$y=-x^2$

for the corresponding chart

2