Locate the Vertex of the Parabola: Analyze y = x² - 6

Q: Why is the vertex at (0, -6) and not (0, 0)?

The y-intercept is where the parabola crosses the y-axis, which happens at $ (0, -6) $. Since this parabola has no linear term (b = 0), the vertex occurs at the same x-coordinate as the y-intercept!

Q: How do I know if this parabola opens up or down?

Look at the coefficient of $ x^2 $! Since a = 1 (positive), this parabola opens upward, making $ (0, -6) $ the minimum point.

Q: What if the equation was y = x² + 6 instead?

Then the vertex would be at $ (0, 6) $! The constant term determines the y-coordinate of the vertex when there's no linear term.

Q: Can I use completing the square for this problem?

You could, but it's unnecessary here! Since $ y = x^2 - 6 $ is already in the form $ y = a(x - h)^2 + k $ where h = 0 and k = -6.

Q: How do I check if my vertex is correct?

Substitute the x-coordinate into the original equation: $ y = (0)^2 - 6 = -6 $. If you get the y-coordinate of your vertex, you're right! ✓

Quadratic Functions with Vertex Form Analysis

Find the vertex of the parabola

$y=x^2-6$

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Find the vertex of the parabola

00:03 Use the formula to describe the parabola function

00:09 The coordinates of the vertex are (P,K)

00:12 Use this formula and find the vertex point

00:15 Write our function as a template of the formula

00:21 Substitute appropriate values according to the given data

00:28 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Find the vertex of the parabola

$y=x^2-6$

Step-by-step solution

To solve this problem, we'll follow these steps:

Step 1: Identify the given information.
Step 2: Apply the vertex formula to find $h$ and $k$ .
Step 3: Determine the coordinates of the vertex.

Now, let's work through each step:
Step 1: The given quadratic equation is $y = x^2 - 6$ where $a = 1$ , $b = 0$ , and $c = -6$ .
Step 2: We use the vertex formula:

$h = -\frac{b}{2a}$
Substituting the values, $h = -\frac{0}{2 \cdot 1} = 0$ .

$k = c - \frac{b^2}{4a}$
Using the given values, $k = -6 - \frac{0^2}{4 \cdot 1} = -6$ .

Step 3: Therefore, the vertex of the parabola is at $(h, k) = (0, -6)$ .

The solution to the problem is $(0, -6)$ .

Final Answer

$(0,-6)$

Key Points to Remember

Essential concepts to master this topic

Standard Form: For $y = ax^2 + bx + c$ , identify coefficients a, b, c
Vertex Formula: Use $h = -\frac{b}{2a}$ and $k = f(h)$ to find coordinates
Verification: Substitute vertex coordinates back: $y = 0^2 - 6 = -6$ ✓

Common Mistakes

Avoid these frequent errors

Using wrong vertex formula for k-value
Don't use $k = c - \frac{b^2}{4a}$ when you can substitute directly = unnecessary complexity! When b = 0, just substitute h into the original equation. Always use $k = f(h)$ by plugging h-value back into $y = x^2 - 6$ .

Practice Quiz

Test your knowledge with interactive questions

The following function has been graphed below:

$$ f(x)=x^2-6x $$

Calculate point C.

FAQ

Everything you need to know about this question

Why is the vertex at (0, -6) and not (0, 0)?

The y-intercept is where the parabola crosses the y-axis, which happens at $(0, -6)$ . Since this parabola has no linear term (b = 0), the vertex occurs at the same x-coordinate as the y-intercept!

How do I know if this parabola opens up or down?

Look at the coefficient of $x^2$ ! Since a = 1 (positive), this parabola opens upward, making $(0, -6)$ the minimum point.

What if the equation was y = x² + 6 instead?

Then the vertex would be at $(0, 6)$ ! The constant term determines the y-coordinate of the vertex when there's no linear term.

Can I use completing the square for this problem?

You could, but it's unnecessary here! Since $y = x^2 - 6$ is already in the form $y = a(x - h)^2 + k$ where h = 0 and k = -6.

How do I check if my vertex is correct?

Substitute the x-coordinate into the original equation: $y = (0)^2 - 6 = -6$ . If you get the y-coordinate of your vertex, you're right! ✓

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Locate the Vertex of the Parabola: Analyze y = x² - 6

Quadratic Functions with Vertex Form Analysis

❤️ Continue Your Math Journey!

Step-by-step video solution

Step-by-step written solution

Understand the problem

Step-by-step solution

Final Answer

Key Points to Remember

Common Mistakes

Practice Quiz

FAQ

Why is the vertex at (0, -6) and not (0, 0)?

How do I know if this parabola opens up or down?

What if the equation was y = x² + 6 instead?

Can I use completing the square for this problem?

How do I check if my vertex is correct?

🌟 Unlock Your Math Potential

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