A ball is thrown straight up from a height of 3 ft with a speed of [tex]$32 \, \text{ft/s}$[/tex]. Its height above the ground after [tex]$x$[/tex] seconds is given by the quadratic function [tex]y = -16x^2 + 32x + 3[/tex].

Explain the steps you would use to determine the path of the ball in terms of a transformation of the graph of [tex]y = x^2[/tex].



Answer :

To analyze the path of the ball and describe it in terms of transformations of the graph of [tex]\( y = x^2 \)[/tex], follow these steps:

1. Understand the given quadratic function:
- The height [tex]\( y \)[/tex] of the ball at any time [tex]\( x \)[/tex] seconds is given by the quadratic function [tex]\( y = -16x^2 + 32x + 3 \)[/tex].

2. Identify the standard form:
- The given function is in the standard quadratic form [tex]\( y = ax^2 + bx + c \)[/tex].

3. Determine the coefficients:
- [tex]\( a = -16 \)[/tex] (coefficient of [tex]\( x^2 \)[/tex])
- [tex]\( b = 32 \)[/tex] (coefficient of [tex]\( x \)[/tex])
- [tex]\( c = 3 \)[/tex] (constant term)

4. Calculate the vertex of the parabola:
- The vertex form of a quadratic function [tex]\( y = a(x-h)^2 + k \)[/tex] provides the vertex [tex]\((h, k)\)[/tex].
- To find the vertex [tex]\((h, k)\)[/tex], use the formulas:
[tex]\[ h = -\frac{b}{2a} \][/tex]
[tex]\[ k = f(h) \][/tex]
- Plug in the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[ h = -\frac{32}{2(-16)} = 1 \][/tex]
- To find [tex]\( k \)[/tex], substitute [tex]\( h \)[/tex] back into the original function:
[tex]\[ k = -16(1)^2 + 32(1) + 3 = -16 + 32 + 3 = 19 \][/tex]
- So, the vertex of the parabola is [tex]\( (1, 19) \)[/tex].

5. Interpret the transformations:
- The standard quadratic function [tex]\( y = x^2 \)[/tex] undergoes several transformations to become the given function [tex]\( y = -16x^2 + 32x + 3 \)[/tex]:
- Reflection: The negative sign of [tex]\( a \)[/tex] (i.e., -16) indicates that the parabola [tex]\( y = -16x^2 \)[/tex] is reflected over the x-axis as compared to [tex]\( y = x^2 \)[/tex].
- Vertical Stretch: The coefficient [tex]\( |a| = 16 \)[/tex] shows that the parabola is stretched vertically by a factor of 16.
- Horizontal Shift: The vertex [tex]\( h = 1 \)[/tex] indicates that the parabola is shifted horizontally to the right by 1 unit.
- Vertical Shift: The vertex [tex]\( k = 19 \)[/tex] indicates that the parabola is shifted vertically up by 19 units.

6. Summary of transformations:
- Starting from the graph of [tex]\( y = x^2 \)[/tex]:
- Reflect it over the x-axis.
- Stretch it vertically by a factor of 16.
- Shift it horizontally to the right by 1 unit.
- Shift it vertically upward by 19 units.

The path of the ball corresponds to these transformations, starting from the base graph of [tex]\( y = x^2 \)[/tex]. The vertex [tex]\((1, 19)\)[/tex] shows the maximum height reached by the ball at 1 second, and the coefficients dictate the shape and direction of the parabola.