Find the Quadratic Function with Minimum Point (-2,0): Parabola Identification

Q: Why is the vertex form y = (x - h)² + k instead of y = (x + h)² + k?

The vertex form uses subtraction because when $x = h$, the expression $(x - h)$ equals zero, making that the vertex's x-coordinate. It's the standard mathematical convention!

Q: How do I handle negative coordinates in the vertex?

When the vertex has a negative x-coordinate like $(-2, 0)$, substitute carefully: $x - (-2) = x + 2$. The double negative becomes positive!

Q: What if my parabola opens downward?

Add a negative coefficient in front: $y = -(x - h)^2 + k$. The vertex form still works the same way, but the parabola has a maximum point instead of minimum.

Q: Can I expand the vertex form to standard form?

Yes! For $y = (x + 2)^2$, expand to get $y = x^2 + 4x + 4$. Both forms represent the same parabola, but vertex form shows the vertex directly.

Quadratic Functions with Vertex Form

Which function corresponds to a parabola with a minimum point of $(-2,0)$ ?

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Find the appropriate function for the parabola with the minimum point

00:03 A smiling parabola has a positive coefficient for X squared

00:06 In this parabola, the intersection point is at the origin

00:10 We need a sad parabola 2 steps to the left

00:21 Negative times negative always equals positive

00:27 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

Which function corresponds to a parabola with a minimum point of $(-2,0)$ ?

Step-by-step solution

To solve the problem, we need to write the equation of a parabola with the given vertex.

Step 1: Identify the form of the equation. For a parabola with vertex $(h, k)$ , the equation is $y = (x - h)^2 + k$ .

Step 2: Plug in the coordinates of the vertex. Here, the vertex is $(-2, 0)$ , so $h = -2$ and $k = 0$ .

Step 3: Substitute into the vertex form:

Replace $h = -2$ and $k = 0$ into the equation:
$y = (x - (-2))^2 + 0$

Step 4: Simplify the equation.

This results in:

$y = (x + 2)^2$ .

Therefore, the function corresponding to the given parabola is $y = (x + 2)^2$ .

Final Answer

$y=(x+2)^2$

Key Points to Remember

Essential concepts to master this topic

Vertex Form: For vertex $(h, k)$ , equation is $y = (x - h)^2 + k$
Sign Rule: For vertex $(-2, 0)$ , get $x - (-2) = x + 2$
Check: Substitute $x = -2$ : $(-2 + 2)^2 = 0$ ✓

Common Mistakes

Avoid these frequent errors

Confusing signs when substituting negative h-values
Don't write y = (x - (-2))^2 as y = (x - 2)^2 = wrong vertex at (2, 0)! The double negative becomes positive. Always simplify x - (-2) = x + 2 for vertex (-2, 0).

Practice Quiz

Test your knowledge with interactive questions

Find the intersection of the function

$$ y=(x+4)^2 $$

With the Y

FAQ

Everything you need to know about this question

Why is the vertex form y = (x - h)² + k instead of y = (x + h)² + k?

The vertex form uses subtraction because when $x = h$ , the expression $(x - h)$ equals zero, making that the vertex's x-coordinate. It's the standard mathematical convention!

How do I handle negative coordinates in the vertex?

When the vertex has a negative x-coordinate like $(-2, 0)$ , substitute carefully: $x - (-2) = x + 2$ . The double negative becomes positive!

What if my parabola opens downward?

Add a negative coefficient in front: $y = -(x - h)^2 + k$ . The vertex form still works the same way, but the parabola has a maximum point instead of minimum.

Can I expand the vertex form to standard form?

Yes! For $y = (x + 2)^2$ , expand to get $y = x^2 + 4x + 4$ . Both forms represent the same parabola, but vertex form shows the vertex directly.

How do I verify my answer is correct?

Substitute the vertex coordinates into your equation. For vertex $(-2, 0)$ and $y = (x + 2)^2$ : when $x = -2$ , $y = (-2 + 2)^2 = 0$ ✓

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