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Graph y > x^2 - 5. Click on the graph until the correct one appears.

Graph y > x^2 - 5. Click on the graph until the correct one appears.-example-1
Graph y > x^2 - 5. Click on the graph until the correct one appears.-example-1
Graph y > x^2 - 5. Click on the graph until the correct one appears.-example-2
Graph y > x^2 - 5. Click on the graph until the correct one appears.-example-3
Graph y > x^2 - 5. Click on the graph until the correct one appears.-example-4

2 Answers

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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)

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


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

Graph y > x^2 - 5. Click on the graph until the correct one appears.-example-1
Graph y > x^2 - 5. Click on the graph until the correct one appears.-example-2
User Vincent Roye
by
7.9k points
0 votes

Answer:

Graph 2

Explanation:

Here, the given inequality is,


y>x^2-5-----(1)

At (0,0),


0>0-5


\implies 0 > -5 ( True )

So, The shaded region must contain the origin,

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

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


y=a(x-h)^2+k

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

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


y=(x-0)^2+(-5)

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.

Graph y > x^2 - 5. Click on the graph until the correct one appears.-example-1
User Prashant Bhanarkar
by
8.0k points

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