Chapter 9.1, Problem 21CYU

(a)

To determine

To Graph:

The equation.

Explanation of Solution

Given Information:

A juggler is tossing a ball into the air.The height of the ball in feet can be modeled by the equation $y=-16x^2+16x+5$ , where y represents the height of the ball at x seconds.

Concept Used :

• Quadratic Functions:
• Quadratic functions are non-linear and are of the form:

  $f(x)=a{x}^2+bx+c;a\ne 0$

This is the standard form of Quadratic function.

• Shape of the graph of a Quadratic function is called Parabola.
• Parabolas are symmetric about a central line called Axis Of Symmetry.
• The axis of symmetry intersects a parabola only at one point , called the Vertex.
• Parent Function:

  $y=x^2$

• Standard Form :

  $f(x)=a{x}^2+bx+c;a\ne 0$

• Type of Graph:

Parabola

• Axis of Symmetry: Passes through the vertex and divides the parabola into two congruent halves.

  $x=-\frac{b}{2a}$

• y-intercept = The y coordinate of the point at which the graph intersects y-axis = c
• Graph:

  

• If a > 0 ,
• The graph of $a{x}^2+bx+c$ opens upward.
• The lowest point on the graph is the minimum.
• If a < 0 ,
• The graph of $a{x}^2+bx+c$ opens downward.
• The highest point on the graph is the maximum.
• The maximum or the minimum is the vertex.

Calculation:

• In the equation $y=-16x^2+16x+5$ ; comparing it with standard form of quadratic equation , we have a = -16 , b = 16 , c = 5 .
• Find equation of Axis of Symmetry :

  $\begin{array}{l}y=a{x}^2+bx+c\mathrm{...........................}[s\tan dard\_form]\\ x=-\frac{b}{2a}\mathrm{.....................................}[formula\_of\_axis\_of\_symmetry]\\ x=-\frac{16}{2(-16)}=\frac{1}{2}\mathrm{..........................}[a=-16,b=16]\end{array}$

The equation of axis of symmetry is $x=\frac{1}{2}$

• Find the Vertex and determine whether it is maximum or minimum:

Since axis of symmetry passes through the vertex , we have x coordinate of the vertex (x,y) is $x=\frac{1}{2}$

Substituting the value of $x=\frac{1}{2}$in the equation $y=-16x^2+16x+5$

  $\begin{array}{l}y=-16x^2+16x+5\\ $ y=-16{(\frac{1}{2})}^2+16(\frac{1}{2})+\mathrm{5...............}[x=\frac{1}{2}]$y=-4+8+\mathrm{5.................................}[simplify]\\ y=9\end{array}$

The vertex is at $(\frac{1}{2},9)$ .

Since the a = -16 < 0 .So, the graph opens downward .Hence, the vertex is the maximum.

• Find y-intercept :

The y-coordinate of the point at which the graph cuts the y-axis is the y-intercept.

y-intercept always occurs at (0,c) and c = 5 here.

So, y-intercept = 5 and the point is (0,5)

• The axis of symmetry cuts the parabola into two equal parts . So, if there is a point on one side, there is a corresponding point on the other side that is at a same distance from the axis of symmetry and the same y-value.

x	y
0	5
$\frac{1}{2}$	9
1	5

• Connect the points with a smooth curve.

Graph :

  

(b)

To determine

To Find:

At what height is the ball thrown.

Answer to Problem 21CYU

5 feet

Explanation of Solution

Given Information:

A juggler is tossing a ball into the air.The height of the ball in feet can be modeled by the equation $y=-16x^2+16x+5$ , where y represents the height of the ball at x seconds.

Concept Used :

• Quadratic Functions:
• Quadratic functions are non-linear and are of the form:

  $f(x)=a{x}^2+bx+c;a\ne 0$

This is the standard form of Quadratic function.

• y-intercept = The y coordinate of the point at which the graph intersects y-axis = c

Calculation:

• In the equation $y=-16x^2+16x+5$ ; comparing it with standard form of quadratic equation , we have a = -16 , b = 16 , c = 5 .
• To find the height of ball initially, we have time = 0 second :

That is , the y-intercept of the quadratic equation.

  $\begin{array}{l}y=-16x^2+16x+5\\ $ y=-16{(0)}^2+16(0)+\mathrm{5....................}[x=0]$y=5\end{array}$

So, initially , the ball is thrown at a height of 5 feet.

(c)

To determine

To Find:

The maximum height of the ball.

Answer to Problem 21CYU

9 feet

Explanation of Solution

Given Information:

From part (a) , we have the vertex is $(\frac{1}{2},9)$ .

Concept Used :

• Quadratic Functions:
• Quadratic functions are non-linear and are of the form:

  $f(x)=a{x}^2+bx+c;a\ne 0$

This is the standard form of Quadratic function.

• Shape of the graph of a Quadratic function is called Parabola.
• Parabolas are symmetric about a central line called Axis Of Symmetry.
• The axis of symmetry intersects a parabola only at one point , called the Vertex.
• If a > 0 ,
• The graph of $a{x}^2+bx+c$ opens upward.
• The lowest point on the graph is the minimum.
• If a < 0 ,
• The graph of $a{x}^2+bx+c$ opens downward.
• The highest point on the graph is the maximum.
• The maximum or the minimum is the vertex.

Calculation:

• In the equation $y=-16x^2+16x+5$ ; comparing it with standard form of quadratic equation , we have a = -16 , b = 16 , c = 5 .
• The Vertex and determine whether it is maximum or minimum:

The vertex is at $(\frac{1}{2},9)$ .[From part (a)

Since the a = -16 < 0 .So, the graph opens downward .Hence, the vertex is the maximum.

• The maximum height occurs at vertex.

So, the maximum height the ball reaches is 9 feet after $\frac{1}{2}$ seconds of release.