Given the function:$$ y=x^2 $$Is there a point for ? $$ y=4 $$?

$$ (2,-4) $$$$ (-2,-4) $$

$$ (2,8) $$$$ (-2,-8) $$

The functions y=x²

The functions $(y=x^2,y=-x^2,y=ax^2 )$

$Y=X^2$

Properties of the function:

The most basic quadratic function $b=0$ , $c=0$
Minimum, happy face function, its vertex is $(0,0)$
The axis of symmetry of this function is $X=0$ .
The function's interval of increase: $X>0$
The function's interval of decrease: $X<0$
Set of positivity: Every $X$ except $0$ .
Set of negativity: None. The entire parabola is above the axis $X$ .

Detailed explanation

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Test yourself on parabola of the form y=x²!

What is the value of y for the function?

$$ y=x^2 $$

of the point $$ x=2 $$ ?

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$y=-x^2$

Properties of the Function

The most basic quadratic function $a=-1$ , $b=0$ , $c=0$
Maximum, sad face function, its vertex is $(0,0)$
The axis of symmetry of this function is $X=0$ .
The interval of increase of the function: $X<0$
The interval of decrease of the function: $X>0$
Set of positivity: None. The entire parabola is below the axis $X$ .
Set of negativity: All $X$ except for $X=0$

$y=ax^2$

Properties of the function:
The quadratic function
$any number = a$ , $b=0$ , $c=0$

Its vertex is $(0,0)$
The axis of symmetry of this function is $X=0$ .

As $a$ increases, the parabola will have a smaller opening - closer to its axis of symmetry.
As $a$ decreases, the parabola will have a larger opening - further from its axis of symmetry.