Respuesta :

gmany

[tex]y=x^2\\\\for\ x=\pm2\to y=(\pm2)^2=4\to(-2;\ 4);\ (2;\ 4)\\\\for\ x=\pm1\to y=(\pm1)^2=1\to (-1;\ 1);\ (1;\ 1)\\\\for\ x=0\to y=0^2=0\to (0;\ 0)[/tex]

Shift the graph of the function y = x², 5 units down /look at the picture #1/.

[tex]y > x^2-5[/tex] /look at the picture #2/ - your answer

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Answer:

Graph 2

Step-by-step explanation:

Here, the given inequality is,

[tex]y>x^2-5-----(1)[/tex]

At (0,0),

[tex]0>0-5[/tex]

[tex]\implies 0 > -5[/tex]       ( True )

So, The shaded region must contain the origin,

Hence, Graph 1 and Graph 3 can not be the graph of [tex]y>x^2-5[/tex],

Now, we know that, The standard form of a parabola is,

[tex]y=a(x-h)^2+k[/tex]

Where, (h,k) is the vertex of the parabola,

From equation (1), the related equation of inequality (1) can be written,

[tex]y=(x-0)^2+(-5)[/tex]

By Comparing,

The vertex of the related parabola of inequality (1) is (0,-5),

⇒ The vertex of the given parabola must be lie on the negative y-axis

Thus, Graph 3 can not be the graph of the given equation,

Therefore, Graph 2 must be the graph of the given inequality.

Ver imagen parmesanchilliwack

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