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A ball is thrown straight up from a height of 3 ft with a speed of 32ft/s. It’s height above the ground after x seconds is given by the quadratic function y=-16x^2+32x+3
Explain the steps you would use to determine the path of the ball in terms of a transformation of the graph of y=X^2.

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Answer:

My first steps is to enter the equation into desmos and see what the coordinates are which is (1, 19) 1 to the left and 19 up. The function is narrow and goes below the x axis.

Step-by-step explanation: Got it right (btw you can copy my answer and put it on edge) that way our answers aren't all the same :) bye!!

MrRoyal

Transformation involves changing the position of a function.

The path of the ball from y = x^2 is obtained by

  • Shifting the graph of y = x^2 right by 1 unit
  • Then the graph is vertically stretched by -16
  • Lastly, the graph is shifted up by 19 units

The function is given as:

[tex]\mathbf{y=-16x^2+32x+3}[/tex]

Factor out -16

[tex]\mathbf{y=-16(x^2-2x)+3}[/tex]

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Take the coefficient of x

[tex]\mathbf{k = -2}[/tex]

Divide by 2

[tex]\mathbf{k/2 = -1}[/tex]

Take it square

[tex]\mathbf{(k/2)^2 = 1}[/tex]

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Add and subtract the 1 in the bracket of [tex]\mathbf{y=-16(x^2-2x)+3}[/tex]

[tex]\mathbf{y=-16(x^2-2x+ 1 - 1)+3}[/tex]

Open bracket

[tex]\mathbf{y=-16(x^2-2x+ 1) + 16+3}[/tex]

[tex]\mathbf{y=-16(x^2-2x+ 1) + 19}[/tex]

Express as squares

[tex]\mathbf{y=-16(x- 1)^2 + 19}[/tex]

The above means that:

  • The graph of y = x^2 is shifted right by 1 unit
  • Then the graph is vertically stretched by -16
  • Lastly, the graph is shifted up by 19 units

Read more about transformation at:

https://brainly.com/question/11709244

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