graph the equation shown below by transforming the given graph of the parent function. \n$y = 4x^2$

Answer

Explanation:

Step1: Identify the parent function

The parent function for ( y = 4x^2 ) is ( y = x^2 ), which is a parabola opening upwards with vertex at the origin ((0,0)).

Step2: Analyze the transformation

The equation ( y = 4x^2 ) is a vertical stretch of the parent function ( y = x^2 ) by a factor of 4. For a vertical stretch by a factor ( a ) (where ( a>1 )), the graph of ( y = ax^2 ) is narrower than the graph of ( y = x^2 ).

Step3: Determine key points

For the parent function ( y = x^2 ), some key points are ((-2, 4)), ((-1, 1)), ((0, 0)), ((1, 1)), ((2, 4)).
For ( y = 4x^2 ), we multiply the ( y )-coordinates of these points by 4:
- When ( x = -2 ), ( y = 4\times(-2)^2 = 4\times4 = 16 )? Wait, no, wait. Wait, looking at the given graph, the points on the parent graph (probably ( y = x^2 ) scaled? Wait, no, the given graph has points like at ( x = 1 ), ( y = 1 ); ( x = 2 ), ( y = 4 ); ( x = 3 ), ( y = 9 )? Wait, no, the graph in the image has points at ( x = -1 ), ( y = 1 ); ( x = 1 ), ( y = 1 ); ( x = -2 ), ( y = 4 ); ( x = 2 ), ( y = 4 ); ( x = -3 ), ( y = 9 ); ( x = 3 ), ( y = 9 ). Wait, maybe the parent graph here is ( y = x^2 ), and we need to transform it to ( y = 4x^2 ). So for ( y = 4x^2 ), when ( x = 1 ), ( y = 4\times1^2 = 4 ); ( x = 2 ), ( y = 4\times4 = 16 )? No, that can't be. Wait, maybe the given graph is of ( y = x^2 ), and we need to stretch it vertically by 4. Wait, but the graph in the image has points at ( x = 1 ), ( y = 1 ); ( x = 2 ), ( y = 4 ); ( x = 3 ), ( y = 9 ). So to get ( y = 4x^2 ), we need to take each point ((x, y)) on the parent graph (which is ( y = x^2 )) and multiply the ( y )-coordinate by 4. So:
- At ( x = -1 ), original ( y = 1 ), new ( y = 4\times1 = 4 )
- At ( x = 1 ), original ( y = 1 ), new ( y = 4\times1 = 4 )
- At ( x = -2 ), original ( y = 4 ), new ( y = 4\times4 = 16 )? But the graph in the image doesn't go that high. Wait, maybe the given graph is of ( y = x^2 ) with a different scale? Wait, the problem says "transforming the given graph of the parent function". So the given graph is the parent function (probably ( y = x^2 ) or a scaled version). To graph ( y = 4x^2 ), we apply a vertical stretch by 4. So the vertex remains at ((0,0)), and the parabola becomes narrower. So the key points for ( y = 4x^2 ) would be:
  - ( x = -1 ), ( y = 4\times(-1)^2 = 4 )
  - ( x = 1 ), ( y = 4\times1^2 = 4 )
  - ( x = -2 ), ( y = 4\times(-2)^2 = 16 ) (but that's outside the given graph's range)
  - Wait, maybe the given graph is of ( y = x^2 ) with points at ( x = -1, 1 ) (y=1), ( x = -2, 2 ) (y=4), ( x = -3, 3 ) (y=9). So to get ( y = 4x^2 ), we need to take each of these points and multiply the y-coordinate by 4. So:
    - For ( x = -1 ), new y = 4*1 = 4
    - For ( x = 1 ), new y = 4*1 = 4
    - For ( x = -2 ), new y = 4*4 = 16
    - For ( x = 2 ), new y = 4*4 = 16
    - For ( x = -3 ), new y = 4*9 = 36
    - For ( x = 3 ), new y = 4*9 = 36 But the given graph has a y-axis up to 10, so maybe the parent graph is scaled differently. Wait, perhaps the given graph is of ( y = x^2 ) with a vertical scale, but the problem is to graph ( y = 4x^2 ) by transforming this graph. So the transformation is a vertical stretch by 4, so each point ((x, y)) on the parent graph becomes ((x, 4y)). So if the parent graph has a point at ((1, 1)), the new point is ((1, 4)); at ((2, 4)), new point is ((2, 16)), etc. So the graph of ( y = 4x^2 ) will be narrower than the parent graph, with the vertex at the origin, opening upwards, and passing through points like ((1, 4)), ((2, 16)), etc.

Answer:

To graph ( y = 4x^2 ) from the parent function (likely ( y = x^2 ) as shown in the image), apply a vertical stretch by a factor of 4. This makes the parabola narrower:

The vertex remains at ((0, 0)).
For each point ((x, y)) on the parent graph, the corresponding point on ( y = 4x^2 ) is ((x, 4y)). For example:
- If the parent has ((1, 1)), the new point is ((1, 4)).
- If the parent has ((2, 4)), the new point is ((2, 16)).
- The parabola opens upwards, is narrower than the parent, and has its vertex at the origin.