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graph the equation shown below by transforming the given graph of the p…

Question

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

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 \((0,0)\) and a standard "width" (the coefficient of \( x^2 \) is 1).

Step2: Analyze the transformation (vertical stretch)

The given function is \( y = 4x^2 \). The coefficient \( 4 \) (greater than 1) indicates a vertical stretch of the parent function \( y = x^2 \) by a factor of \( 4 \). For a vertical stretch by a factor of \( a \) (where \( a>1 \)) of the function \( y = f(x) \), the transformed function is \( y = a \cdot f(x) \). This means that for each \( x \)-value, the \( y \)-value of the parent function is multiplied by \( 4 \).

Step3: Determine key points for transformation

  • For the parent function \( y = x^2 \), some key points are:
  • When \( x = -2 \), \( y = (-2)^2 = 4 \)
  • When \( x = -1 \), \( y = (-1)^2 = 1 \)
  • When \( x = 0 \), \( y = 0^2 = 0 \)
  • When \( x = 1 \), \( y = 1^2 = 1 \)
  • When \( x = 2 \), \( y = 2^2 = 4 \)
  • After applying the vertical stretch (multiplying \( y \)-values by 4):
  • When \( x = -2 \), \( y = 4\times4 = 16 \)? Wait, no, wait the graph in the image has points. Wait, maybe the parent graph in the image is \( y = x^2 \)? Wait, no, looking at the graph, when \( x = 1 \), the \( y \)-value is 1? Wait, no, the blue points: at \( x = 3 \), \( y = 9 \); \( x = 2 \), \( y = 4 \); \( x = 1 \), \( y = 1 \). Wait, that's \( y = x^2 \). So to graph \( y = 4x^2 \), we need to take each point \((x, y)\) on \( y = x^2 \) and transform it to \((x, 4y)\).
  • For \( x = -3 \): parent \( y = (-3)^2 = 9 \), transformed \( y = 4\times9 = 36 \)? Wait, no, the graph in the image has a grid, maybe the parent graph shown is \( y = x^2 \), but the problem is to graph \( y = 4x^2 \) by transforming that parent graph. Wait, maybe the parent graph in the image is \( y = x^2 \), and we need to apply the vertical stretch. Let's check the key points:
  • At \( x = -1 \): parent \( y = 1 \), transformed \( y = 4\times1 = 4 \)
  • At \( x = 1 \): parent \( y = 1 \), transformed \( y = 4\times1 = 4 \)
  • At \( x = -2 \): parent \( y = 4 \), transformed \( y = 4\times4 = 16 \)
  • At \( x = 2 \): parent \( y = 4 \), transformed \( y = 4\times4 = 16 \)
  • At \( x = 0 \): parent \( y = 0 \), transformed \( y = 4\times0 = 0 \)
  • So the vertex remains at \((0,0)\) (since multiplying 0 by 4 is still 0), and the parabola becomes narrower (since the \( y \)-values increase faster) because of the vertical stretch by a factor of 4.

Step4: Sketch the transformed graph

Using the transformed key points:

  • Vertex at \((0,0)\)
  • When \( x = \pm1 \), \( y = 4 \)
  • When \( x = \pm2 \), \( y = 16 \) (but looking at the grid, maybe the original graph has a smaller scale? Wait, the given graph in the image has points at \( x = \pm3 \), \( y = 9 \); \( x = \pm2 \), \( y = 4 \); \( x = \pm1 \), \( y = 1 \). So to graph \( y = 4x^2 \), we take each of these points and multiply the \( y \)-coordinate by 4:
  • For \( x = -3 \): original \( y = 9 \), new \( y = 4\times9 = 36 \) (but that's outside the given grid, maybe the intended parent graph is \( y = x^2 \) with the shown points, and we need to scale the \( y \)-values by 4. So the new graph will have the same vertex, but steeper (narrower) because the vertical stretch makes the parabola rise more quickly for each \( x \)-value.

Answer:

To graph \( y = 4x^2 \) from the parent function \( y = x^2 \):

  1. Identify the transformation: Vertical stretch by a factor of \( 4 \) (since \( y = 4x^2 = 4 \cdot x^2 \)).
  2. Transform key points of \( y = x^2 \) (e.g., \((-2, 4) \to (-2, 16)\), \((-1, 1) \to (-1, 4)\), \((0, 0) \to (0, 0)\), \((1, 1) \to (1, 4)\), \((2, 4) \to (2, 16)\)).
  3. Sketch the parabola: Plot the transformed points (vertex at \((0,0)\), narrower than \( y = x^2 \) due to the vertical stretch) and draw a smooth curve through them, opening upwards.

(Note: The graph will have the same vertex \((0,0)\) but be steeper (narrower) than the parent parabola \( y = x^2 \) because the \( y \)-values increase 4 times faster for each \( x \).)