Locate the Vertex: Analyzing y = x²

Q: Why is the vertex of y = x² at the origin?

Since $ y = x^2 $ has no linear term (b = 0) and no constant term (c = 0), the parabola is perfectly centered at the origin. The vertex formula gives us x = -0/(2×1) = 0.

Q: How do I remember the vertex formula?

Think of it as finding the axis of symmetry first! The formula $ x = -\frac{b}{2a} $ gives you the x-coordinate where the parabola is perfectly balanced.

Q: What if I get a different vertex for y = x²?

Double-check your coefficients! For $ y = x^2 $, we have a = 1, b = 0, c = 0. Any other values mean you're looking at a different equation.

Q: Do all parabolas have their vertex at (0,0)?

No! Only $ y = x^2 $ has its vertex at the origin. Most parabolas like $ y = x^2 + 2x + 1 $ have vertices elsewhere.

Q: How can I visualize this vertex?

The vertex is the lowest point of the parabola when it opens upward (like a U-shape). For $ y = x^2 $, this happens right at the origin where x = 0 and y = 0.

Quadratic Functions with Vertex Form Analysis

Find the vertex of the parabola

$y=x^2$

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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 We will use the formula to describe a parabolic function

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

00:18 We will write our function as a template of the formula

00:21 We will use this formula and find the vertex point

00:24 We will 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

$y=x^2$

Step-by-step solution

To find the vertex of the parabola $y = x^2$ , we follow these steps:

Step 1: Identify the coefficients from the equation $y = x^2$ . Here, $a = 1$ , $b = 0$ , and $c = 0$ .
Step 2: Use the vertex formula $x = -\frac{b}{2a}$ to find the x-coordinate of the vertex. Substituting the values, we get:

x = -\frac{0}{2 \times 1} = 0

Therefore, the x-coordinate of the vertex is $0$ .

Step 3: Substitute $x = 0$ back into the equation to find the y-coordinate:

y = (0)^2 = 0

Thus, the y-coordinate of the vertex is also $0$ .

Therefore, the vertex of the parabola $y = x^2$ is $(0,0)$ .

The correct answer choice is: $(0,0)$ .

Final Answer

$(0,0)$

Key Points to Remember

Essential concepts to master this topic

Rule: For $y = ax^2 + bx + c$ , vertex x-coordinate is $-\frac{b}{2a}$
Technique: With $y = x^2$ , we have a=1, b=0, so x = -0/(2×1) = 0
Check: Substitute x=0 back: y = (0)² = 0, confirming vertex is (0,0) ✓

Common Mistakes

Avoid these frequent errors

Confusing vertex with y-intercept
Don't assume the vertex is where the parabola crosses the y-axis = wrong location! While they happen to be the same for y = x², this isn't always true. Always use the vertex formula x = -b/(2a) to find the exact vertex coordinates.

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 of y = x² at the origin?

Since $y = x^2$ has no linear term (b = 0) and no constant term (c = 0), the parabola is perfectly centered at the origin. The vertex formula gives us x = -0/(2×1) = 0.

How do I remember the vertex formula?

Think of it as finding the axis of symmetry first! The formula $x = -\frac{b}{2a}$ gives you the x-coordinate where the parabola is perfectly balanced.

What if I get a different vertex for y = x²?

Double-check your coefficients! For $y = x^2$ , we have a = 1, b = 0, c = 0. Any other values mean you're looking at a different equation.

Do all parabolas have their vertex at (0,0)?

No! Only $y = x^2$ has its vertex at the origin. Most parabolas like $y = x^2 + 2x + 1$ have vertices elsewhere.

How can I visualize this vertex?

The vertex is the lowest point of the parabola when it opens upward (like a U-shape). For $y = x^2$ , this happens right at the origin where x = 0 and y = 0.

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