Question

question
move at least one of the 5 guide points below to complete the graph of ( y = 2(x + 2)^2 ). moving the red points changes the vertical stretch or compression. moving the blue point shifts the function left/right/up/down. click the buttons below to start over or reflect over the x - axis.
(there is a coordinate graph with a parabola, guide points, and buttons like reset, reflect over x - axis)

Explanation:

Step1: Identify the parent function

The given function is \( y = 2(x + 2)^2 \), and the parent function is \( y = x^2 \) (a parabola with vertex at \((0,0)\)).

Step2: Analyze the transformations

• Horizontal shift: The term \((x + 2)\) indicates a shift of the parent function \( y = x^2 \) 2 units to the left (since the form is \( y = a(x - h)^2 + k \), here \( h=-2 \)).
• Vertical stretch: The coefficient \( 2 \) (greater than 1) indicates a vertical stretch of the parent function by a factor of 2.
• Vertex: The vertex of \( y = 2(x + 2)^2 \) is at \((-2, 0)\) (from the vertex form \( y = a(x - h)^2 + k \), where \( h=-2 \), \( k = 0 \)).

Step3: Move the guide points

To graph \( y = 2(x + 2)^2 \):

• Take the vertex of the parent function (0,0) and shift it 2 units left to (-2, 0) (this is the blue point, likely the vertex).
• For other guide points (e.g., for \( y = x^2 \), when \( x = -1 \), \( y = 1 \); \( x = 1 \), \( y = 1 \); \( x = -3 \), \( y = 9 \); \( x = 3 \), \( y = 9 \)), apply the transformations:
• For a point \((x, y)\) on \( y = x^2 \), the corresponding point on \( y = 2(x + 2)^2 \) is found by first shifting \( x \) to \( x - 2 \) (reverse of the horizontal shift) and then multiplying \( y \) by 2 (vertical stretch).
• For example, take the point (1,1) on \( y = x^2 \): shift \( x \) to \( 1 - 2=-1 \), multiply \( y \) by 2: \( y = 2\times1 = 2 \). So the point becomes (-1, 2).
• Similarly, take ( - 1,1) on \( y = x^2 \): shift \( x \) to \( - 1-2=-3 \), multiply \( y \) by 2: \( y = 2\times1 = 2 \). So the point becomes (-3, 2).
• For (2,4) on \( y = x^2 \): shift \( x \) to \( 2 - 2 = 0 \), multiply \( y \) by 2: \( y = 2\times4 = 8 \). Wait, maybe better to use the vertex form directly. For \( x=-2 + 1=-1 \), \( y = 2(1)^2=2 \); \( x=-2 - 1=-3 \), \( y = 2(1)^2 = 2 \); \( x=-2+2 = 0 \), \( y = 2(2)^2=8 \); \( x=-2 - 2=-4 \), \( y = 2(2)^2 = 8 \). So the guide points (red points) should be adjusted accordingly: move the vertex (blue point) to (-2, 0), and other points (e.g., the ones at x = -1, x = 1 relative to the parent vertex) should be shifted left by 2 and stretched vertically by 2.

If we consider the initial graph (with vertex at (0,0) and other points symmetric around the y - axis), to get to \( y = 2(x + 2)^2 \):

• Move the blue vertex point (initially at (0,0)) 2 units to the left to (-2, 0).
• For the red points (e.g., the ones at x = 1 and x=-1, y = 1 on the initial graph), shift them 2 units to the left (so x becomes -1 and -3) and stretch them vertically by a factor of 2 (so y becomes 2 instead of 1). Similarly, for the points at x = 2 and x=-2 (if any), shift x to 0 and -4 and stretch y by 2.

Answer:

To graph \( y = 2(x + 2)^2 \):

1. Move the blue vertex point (originally at \((0,0)\)) 2 units to the left to \((-2, 0)\).
2. For the red guide points:

• Take the points that are 1 unit left/right of the vertex on the initial graph (e.g., \((-1,1)\) and \((1,1)\)): shift them 2 units left (to \((-3,1)\) and \((-1,1)\)) and then stretch them vertically by a factor of 2 (so their y - coordinates become 2, resulting in \((-3,2)\) and \((-1,2)\)).
• Take the points that are 3 units left/right of the vertex on the initial graph (e.g., \((-3,9)\) and \((3,9)\)): shift them 2 units left (to \((-5,9)\) and \((1,9)\)) and then stretch them vertically by a factor of 2 (so their y - coordinates become 18, resulting in \((-5,18)\) and \((1,18)\)). (The specific points to move depend on the initial guide points provided in the graph, but the key is to apply the horizontal shift left by 2 and vertical stretch by 2 to the parent function's points.)