Math Problem Statement

The path of an object hurled from the point (-1,0) is the parabola y = (5/81)(x + 1)(17 - x). Determine the maximum height by finding the vertex.

Solution

The given equation represents the path of a projectile, and we are asked to determine its maximum height by finding the vertex of the parabola.

The equation of the parabola is given as:

y=581(x+1)(17x)y = \frac{5}{81}(x + 1)(17 - x)

Step 1: Expand the quadratic expression

First, we expand the product (x+1)(17x)(x + 1)(17 - x):

(x+1)(17x)=17x2+x(x + 1)(17 - x) = 17 - x^2 + x

So, the equation becomes:

y=581(17x2+x)y = \frac{5}{81}(17 - x^2 + x)

Step 2: Distribute the 581\frac{5}{81}

y=581(17)581(x2)+581(x)y = \frac{5}{81}(17) - \frac{5}{81}(x^2) + \frac{5}{81}(x)

y=8581581x2+581xy = \frac{85}{81} - \frac{5}{81}x^2 + \frac{5}{81}x

Now, the equation is in standard quadratic form:

y=581x2+581x+8581y = -\frac{5}{81}x^2 + \frac{5}{81}x + \frac{85}{81}

Step 3: Find the x-coordinate of the vertex

For a quadratic equation y=ax2+bx+cy = ax^2 + bx + c, the x-coordinate of the vertex is given by:

x=b2ax = -\frac{b}{2a}

Here, a=581a = -\frac{5}{81} and b=581b = \frac{5}{81}, so:

x=5812×581=12x = -\frac{\frac{5}{81}}{2 \times -\frac{5}{81}} = \frac{1}{2}

Step 4: Find the y-coordinate of the vertex

To find the maximum height, we substitute x=12x = \frac{1}{2} into the equation for yy:

y=581(x+1)(17x)y = \frac{5}{81}(x + 1)(17 - x)

Substitute x=12x = \frac{1}{2}:

y=581(12+1)(1712)y = \frac{5}{81}\left(\frac{1}{2} + 1\right)\left(17 - \frac{1}{2}\right)

y=581×32×332y = \frac{5}{81} \times \frac{3}{2} \times \frac{33}{2}

y=581×994=495324=5536y = \frac{5}{81} \times \frac{99}{4} = \frac{495}{324} = \frac{55}{36}

Thus, the maximum height is:

y=55361.53units.y = \frac{55}{36} \approx 1.53 \, \text{units}.

Conclusion:

The maximum height is approximately 5536\frac{55}{36} units.

Would you like further details or have any questions?


Here are 5 relative questions to explore further:

  1. How do we derive the vertex formula for a parabola?
  2. What is the significance of the vertex in projectile motion?
  3. How does changing the coefficient 581\frac{5}{81} affect the parabola's shape?
  4. Can we find the total distance the object traveled horizontally?
  5. How does the starting point (1,0)(-1,0) influence the equation?

Tip: In a quadratic equation, the vertex represents the maximum or minimum value, depending on the sign of the leading coefficient aa.

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Math Problem Analysis

Mathematical Concepts

Quadratic Functions
Vertex of a Parabola
Projectile Motion

Formulas

Vertex formula for a parabola x = -b/(2a)
Quadratic expansion (x + 1)(17 - x)

Theorems

Vertex Form of a Quadratic Function

Suitable Grade Level

High School