Questions: My LTC - Lanier Technical College My Apps Dashboard Lanier Technical C... Applications of Maximum or Minimum The height y (in feet) of a ball thrown by a child is y = -(1/12) x^2 + 6x + 3 where x is the horizontal distance in feet from the point at which the ball is thrown. (a) How high is the ball when it leaves the child's hand? feet (b) What is the maximum height of the ball? feet (c) How far from the child does the ball strike the ground? (Round to the nearest foot.)

My LTC - Lanier Technical College
My Apps Dashboard Lanier Technical C...

Applications of Maximum or Minimum

The height y (in feet) of a ball thrown by a child is
y = -(1/12) x^2 + 6x + 3
where x is the horizontal distance in feet from the point at which the ball is thrown.
(a) How high is the ball when it leaves the child's hand? feet
(b) What is the maximum height of the ball? feet
(c) How far from the child does the ball strike the ground? (Round to the nearest foot.)

Transcript text: My LTC - Lanier Technical College My Apps Dashboard | Lanier Technical C... Applications of Maximum or Minimum The height $y$ (in feet) of a ball thrown by a child is \[ y=-\frac{1}{12} x^{2}+6 x+3 \] where $x$ is the horizontal distance in feet from the point at which the ball is thrown. (a) How high is the ball when it leaves the child's hand? $\square$ feet (b) What is the maximum height of the ball? $\square$ feet (c) How far from the child does the ball strike the ground? (Round to the nearest foot.) $\square$ Question Help: Video Message instructor Submit Question

Solution

Solution Steps

To solve the given problem, we need to analyze the quadratic equation representing the height of the ball as a function of the horizontal distance.

(a) To find the height of the ball when it leaves the child's hand, we evaluate the function at $ x = 0 $.

(b) To find the maximum height of the ball, we need to determine the vertex of the parabola. The x-coordinate of the vertex for a quadratic equation $ ax^2 + bx + c $ is given by $ x = -\frac{b}{2a} $. We then substitute this x-coordinate back into the equation to find the maximum height.

(c) To find the distance from the child where the ball strikes the ground, we need to solve for $ x $ when $ y = 0 $. This involves solving the quadratic equation $ -\frac{1}{12} x^2 + 6x + 3 = 0 $.

Step 1: Height of the Ball When It Leaves the Child's Hand

To find the height of the ball when it leaves the child's hand, we evaluate the function at $ x = 0 $: \[ y = -\frac{1}{12} \cdot 0^2 + 6 \cdot 0 + 3 = 3 \] Thus, the height of the ball when it leaves the child's hand is $ 3 $ feet.

Step 2: Maximum Height of the Ball

The maximum height of the ball occurs at the vertex of the parabola. The x-coordinate of the vertex for the quadratic equation $ y = -\frac{1}{12} x^2 + 6x + 3 $ is given by: \[ x = -\frac{b}{2a} = -\frac{6}{2 \cdot -\frac{1}{12}} = 36 \] Substituting $ x = 36 $ back into the equation to find the maximum height: \[ y = -\frac{1}{12} \cdot 36^2 + 6 \cdot 36 + 3 = 111 \] Thus, the maximum height of the ball is $ 111 $ feet.

Step 3: Distance from the Child Where the Ball Strikes the Ground

To find the distance from the child where the ball strikes the ground, we solve for $ x $ when $ y = 0 $: \[ -\frac{1}{12} x^2 + 6x + 3 = 0 \] Solving this quadratic equation, we get: \[ x \approx 72.4966 \] Thus, the ball strikes the ground approximately $ 72.4966 $ feet from the child.