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Vertex and X-Intercepts of Quadratics

Vertex and X-Intercepts of Quadratics

Assessment

Interactive Video

Mathematics

9th - 10th Grade

Practice Problem

Hard

Created by

Thomas White

FREE Resource

The video tutorial explains how to solve a quadratic equation by finding its vertex and X-intercepts. It begins with an introduction to the problem and the quadratic equation in standard form. The instructor then explains how to identify the vertex using the vertex form of the equation. The process of finding the X-intercepts is discussed, highlighting the issue of no real roots in this particular problem. The tutorial concludes with a note on imaginary roots, which will be covered in future lessons.

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9 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the given quadratic function in the problem?

f(x) = x^2 - 2x + 3

f(x) = x^2 - 4x + 4

f(x) = x - 2^2 + 3

f(x) = x^2 + 2x + 3

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

In the vertex form of a quadratic equation, what does the vertex represent?

The highest point of the graph

The point where the graph crosses the y-axis

The turning point of the graph

The point where the graph crosses the x-axis

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the vertex of the quadratic function f(x) = (x - 6)^2 + 3?

(6, 3)

(-6, 3)

(3, 6)

(6, -3)

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Why do some students mistakenly identify the vertex as (-6, 3)?

They forget to square the x term

They confuse the signs inside the parentheses

They use the wrong formula

They miscalculate the constant term

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the first step in finding the X-intercepts of a quadratic function?

Differentiate the function

Find the vertex

Complete the square

Set f(x) to zero

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What equation do you solve to find the X-intercepts of f(x) = (x - 6)^2 + 3?

0 = x^2 + 6x + 3

0 = x^2 - 6x + 3

0 = (x + 6)^2 - 3

0 = (x - 6)^2 + 3

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Why can't we find real roots for the equation 0 = (x - 6)^2 + 3?

The equation is already solved

The equation has no solutions

The square root of a negative number is not real

The equation is not quadratic

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