In this section, you will write a verbal description of the given quadratic equation.

A verbal description of a quadratic functions graph mentions

• the location of any zeros, roots, or x-intercepts;



• the type of zeros¸ roots, or x-intercepts;

✔Possible roots include: real, rational, irrational or imaginary numbers.

✔ Remember you must actually know the numerical value to be able to decide whether that number is rational or irrational.

✔ Study the graphs below to determine from a graph whether a parabola has real or imaginary roots.

• the location of the y-intercept;



• the location of the vertex;
• whether the parabola has a minimum or maximum;



• the location of the axis of symmetry;



• and where the parabola is increasing or decreasing.

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Review

When you need to write a verbal description for a quadratic function using its graph, you can identify its critical attributes and behavior which include:

• the location of any zeros, roots, or x-intercepts,
• the type of zeros, roots, or x-intercepts,
• the location of the y-intercept,
• the location of the vertex and axis of symmetry
• whether the parabola opens up or down
• whether the parabola has a minimum or maximum
• where the parabola is increasing or decreasing

Example

Write a verbal description for y = 2(x – 3)² – 8

• First find the zeros of the function, set y = 0.

Simplify first then use any of these methods to solve for x:

1. Factoring
2. The Quadratic Formula
3. Completing the Square
4. Graphing

y = 2(x – 3)² – 8
0 = 2(x – 3)² – 8
0 = 2x² – 12x + 10

Solving for x → x = 1 or x = 5, both of which are rational real numbers.

The function has 2 real rational roots and the x-intercepts are at (1, 0) and (5, 0).

The function has a y-intercept at (0, 10).

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More Information

Remember that anytime you square a number it becomes positive, like (3)² = 9 or (- 4)² = 16.

However, if a negative sign is in FRONT of parentheses it is NOT included when squaring and then the answer is negative.
Example: - (5)² = - 25, because it really means – (5 * 5) which is - 25.

• Find the vertex and axis of symmetry.
• Find the vertex. Since the quadratic equation is written in vertex form, the the vertex is located at (h, k).

  y = a(x – h)² + k
  y = 2(x – 3)² – 8
  h = 3 and k = - 8

  The vertex of the parabola is at (3, - 8)

• Find the axis of symmetry. The axis of symmetry passes through the vertex, and its equation is x = the x-coordinate of the vertex → x = 3.
• Since the x² term is positive → the parabola opens up.
• The parabola opens up, it has a minimum.
• The parabola has a minimum, then it must change from decreasing to increasing around the vertex (3, - 8), the graph decreases for x < - 3, then increases for x > - 3.

Extra Practice

Example

While standing on a 10 foot ladder, Annie threw a beach ball into the air. The height in feet of the beach ball can be modeled by the equation f(t) = - 2t² + 8t + 10 where t represents the time in seconds after Annie lets go of the ball and f(t) represents its height above the ground in feet.

1. What was the height of the ball 1 minute 30 seconds after she threw it into the air?
2. How long did it take for the ball to reach the ground?
3. What was the maximum height of the beach ball, and how long did it take to reach that height?

Solution

1. What was the height of the ball 1 minute 30 seconds after she threw it into the air?

   To find the height, substitute the time into the function.
   * (1 minute 30 seconds = 1.5 minutes, so use t = 1.5)

       f (t) = - 2t² + 8t + 10
   f (1.5) = - 2(1.5)² + 8(1.5) +10
   f (1.5) = - 2(2.25) + 12 +10
   f (1.5) = 17.5

   After 1 minute 30 seconds, the beach ball was 17.5 feet above the ground.

2. How long did it take for the ball to reach the ground?

   When the ball touches the ground, its height is 0 → f(t) = 0

   Solve:

   f(t) = -2t² + 8t + 10
   0 = - 2t² + 8t + 10
   0 = - 2 (t² - 4t – 5)
   0 = - 2 (t + 1) (t – 5)
   t = - 1 or 5

   Since t represents time and time cannot be negative, then t = - 1 is an extraneous solution.
   Therefore, t = 5 is the ONLY possible solution.

   The beach ball reached the ground 5 seconds after she released it.

3. What was the maximum height of the beach ball and how long did it take to reach that height?

   Remember: Maximum values are found at the vertex of the parabola.

   For an equation written in standard form, use the formula x = − b over two a b 2a to find the vertex.

   f(t) = - 2t² + 8t + 10 → a = - 2 and b = 8

   x = − b over two a b 2a = − eight over two times negative two (8) 2(-2) = − eight over negative four (8) -4 = −(-2) = 2 → the x–coordinate of the vertex is 2.
   To find the y-coordinate, substitute x = 2 seconds into the equation.

   f(t) = - 2t² + 8t + 10
   f (2) = - 2(2)² + 8(2) + 10
   f (2) = - 8 + 16 + 10
   f (2) = 18 feet
   

   Two seconds after she threw the ball it reached a maximum height of 18 feet.