Question

an engineer uses the equation below to model the height of a cannonball shot straight up from ground level. in the model the height ( y ) (in feet) is a function of ( x ), the number of seconds after the cannonball is shot. y = -16x^2 + 32x complete the parts below. (a) graph the parabola ( y = -16x^2 + 32x ). to do so, plot five points on the parabola: the vertex, two points to the left of the vertex, and two points to the right of the vertex. then click on the graph - a - function button. (b) fill in the blanks below to write an inequality for all the values of ( x ) for which the cannonball is gaining height (going up). then choose the best description of this inequality. inequality: ( square < x < square ) description: ( \bigcirc ) the function ( y = -16x^2 + 32x ) is increasing over these values of ( x ) ( \bigcirc ) the function ( y = -16x^2 + 32x ) is decreasing over these values of ( x ) (c) find the maximum value of ( y ) for the function ( y = -16x^2 + 32x ). then choose the best description of the maximum value of ( y ). maximum value of ( y ): ( square ) description: ( \bigcirc ) the time in seconds when the cannonball hits the ground ( \bigcirc ) the height in feet above the ground the cannonball starts at ( \bigcirc ) the highest elevation in feet the cannonball reaches

Explanation:

Part (a)

To graph the parabola \( y = -16x^2 + 32x \), we first find the vertex. For a quadratic function \( y = ax^2 + bx + c \), the x-coordinate of the vertex is \( x = -\frac{b}{2a} \). Here, \( a = -16 \) and \( b = 32 \), so:

$$ x = -\frac{32}{2(-16)} = 1 $$

Substitute \( x = 1 \) into the function to find the y-coordinate of the vertex:

$$ y = -16(1)^2 + 32(1) = 16 $$

So the vertex is \( (1, 16) \).

Now, find two points to the left of the vertex (e.g., \( x = 0 \) and \( x = -1 \)):

• For \( x = 0 \): \( y = -16(0)^2 + 32(0) = 0 \), so the point is \( (0, 0) \).
• For \( x = -1 \): \( y = -16(-1)^2 + 32(-1) = -16 - 32 = -48 \), so the point is \( (-1, -48) \).

Find two points to the right of the vertex (e.g., \( x = 2 \) and \( x = 3 \)):

• For \( x = 2 \): \( y = -16(2)^2 + 32(2) = -64 + 64 = 0 \), so the point is \( (2, 0) \).
• For \( x = 3 \): \( y = -16(3)^2 + 32(3) = -144 + 96 = -48 \), so the point is \( (3, -48) \).

Plot these points (vertex \( (1, 16) \), left points \( (0, 0) \), \( (-1, -48) \), right points \( (2, 0) \), \( (3, -48) \)) and draw the parabola.

Part (b)

A quadratic function \( y = ax^2 + bx + c \) with \( a < 0 \) (since \( a = -16 < 0 \)) opens downward. The function is increasing to the left of the vertex and decreasing to the right of the vertex. The vertex is at \( x = 1 \), and the function starts at \( x = 0 \) (when the cannonball is shot from the ground). So the cannonball is gaining height (the function is increasing) when \( 0 < x < 1 \).

• Inequality: \( 0 < x < 1 \)
• Description: The function \( y = -16x^2 + 32x \) is increasing over these values of \( x \) (first option).

Part (c)

The maximum value of \( y \) occurs at the vertex of the parabola (since it opens downward). We already found the vertex is \( (1, 16) \), so the maximum value of \( y \) is \( 16 \).

• Maximum value of \( y \): \( 16 \)
• Description: The highest elevation in feet the cannonball reaches (third option).

Final Answers

(a)

Plot the points: Vertex \( (1, 16) \), \( (0, 0) \), \( (-1, -48) \), \( (2, 0) \), \( (3, -48) \) and draw the parabola.

(b)

Inequality: \( \boldsymbol{0} < x < \boldsymbol{1} \)
Description: The function \( y = -16x^2 + 32x \) is increasing over these values of \( x \)

(c)

Maximum value of \( y \): \( \boldsymbol{16} \)
Description: The highest elevation in feet the cannonball reaches

Answer:

Part (a)

To graph the parabola \( y = -16x^2 + 32x \), we first find the vertex. For a quadratic function \( y = ax^2 + bx + c \), the x-coordinate of the vertex is \( x = -\frac{b}{2a} \). Here, \( a = -16 \) and \( b = 32 \), so:

$$ x = -\frac{32}{2(-16)} = 1 $$

Substitute \( x = 1 \) into the function to find the y-coordinate of the vertex:

$$ y = -16(1)^2 + 32(1) = 16 $$

So the vertex is \( (1, 16) \).

Now, find two points to the left of the vertex (e.g., \( x = 0 \) and \( x = -1 \)):

• For \( x = 0 \): \( y = -16(0)^2 + 32(0) = 0 \), so the point is \( (0, 0) \).
• For \( x = -1 \): \( y = -16(-1)^2 + 32(-1) = -16 - 32 = -48 \), so the point is \( (-1, -48) \).

Find two points to the right of the vertex (e.g., \( x = 2 \) and \( x = 3 \)):

• For \( x = 2 \): \( y = -16(2)^2 + 32(2) = -64 + 64 = 0 \), so the point is \( (2, 0) \).
• For \( x = 3 \): \( y = -16(3)^2 + 32(3) = -144 + 96 = -48 \), so the point is \( (3, -48) \).

Plot these points (vertex \( (1, 16) \), left points \( (0, 0) \), \( (-1, -48) \), right points \( (2, 0) \), \( (3, -48) \)) and draw the parabola.

Part (b)

A quadratic function \( y = ax^2 + bx + c \) with \( a < 0 \) (since \( a = -16 < 0 \)) opens downward. The function is increasing to the left of the vertex and decreasing to the right of the vertex. The vertex is at \( x = 1 \), and the function starts at \( x = 0 \) (when the cannonball is shot from the ground). So the cannonball is gaining height (the function is increasing) when \( 0 < x < 1 \).

• Inequality: \( 0 < x < 1 \)
• Description: The function \( y = -16x^2 + 32x \) is increasing over these values of \( x \) (first option).

Part (c)

The maximum value of \( y \) occurs at the vertex of the parabola (since it opens downward). We already found the vertex is \( (1, 16) \), so the maximum value of \( y \) is \( 16 \).

• Maximum value of \( y \): \( 16 \)
• Description: The highest elevation in feet the cannonball reaches (third option).

Final Answers

(a)

Plot the points: Vertex \( (1, 16) \), \( (0, 0) \), \( (-1, -48) \), \( (2, 0) \), \( (3, -48) \) and draw the parabola.

(b)

Inequality: \( \boldsymbol{0} < x < \boldsymbol{1} \)
Description: The function \( y = -16x^2 + 32x \) is increasing over these values of \( x \)

(c)

Maximum value of \( y \): \( \boldsymbol{16} \)
Description: The highest elevation in feet the cannonball reaches