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Maximum Height and Vertex in Quadratics

Maximum Height and Vertex in Quadratics

Assessment

Interactive Video

Physics

9th - 10th Grade

Practice Problem

Hard

Created by

Thomas White

FREE Resource

The video tutorial explains how to determine the maximum height of a ball's path, which forms a downward-opening parabola. The maximum height corresponds to the y-value of the vertex. The tutorial discusses methods to find the vertex, focusing on using the formula x = -B/2A due to the complexity of decimals in the quadratic equation. By substituting the x-value back into the equation, the maximum height is calculated to be 4.43 meters at 0.57 seconds.

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15 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What shape does the path of the ball take when determining maximum height?

Hyperbola

Ellipse

Parabola

Circle

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the significance of the negative 'a' value in the quadratic equation?

It has no significance.

It indicates the parabola opens downwards.

It indicates the parabola is horizontal.

It indicates the parabola opens upwards.

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Which value from the vertex represents the maximum height of the ball?

X-value

Y-value

Z-value

T-value

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Why is the y-value important in determining the maximum height?

It represents the time.

It represents the maximum height.

It represents the speed.

It represents the distance.

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Why is completing the square not used in this scenario?

It is not taught in this course.

It is not accurate.

It involves complex decimals.

It is too simple.

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What does the equation's complexity suggest about the method to use?

Use completing the square

Use a graph

Use the vertex formula

Use direct measurement

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the formula used to find the x-value of the vertex?

x = -2a / B

x = 2a / B

x = -B / 2a

x = B / 2a

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