Transforming y=-x^2: Finding the Parabola Negative Everywhere Except at x=-4

Q: Why is it (x + 4) and not (x - 4) to move left?

This is counterintuitive but crucial! To shift left by 4 units, you replace x with (x + 4). Think of it this way: when x = -4, the expression (x + 4) equals zero, making the vertex at x = -4.

Q: How do I know the parabola opens downward?

Look at the coefficient in front! In $ y = -(x + 4)^2 $, the negative sign means it opens downward. Positive coefficient = opens up, negative coefficient = opens down.

Q: What does 'negative everywhere except at x = -4' mean?

It means the parabola has y-values less than zero for all x-values except x = -4, where y = 0. At the vertex (-4, 0), the parabola touches the x-axis but never goes above it.

Q: Could the answer be y = (x + 4)² instead?

No! That parabola opens upward and is positive everywhere except at x = -4. We need a downward-opening parabola that's negative everywhere except the vertex.

Q: How can I verify my answer is correct?

Test a few points! For $ y = -(x + 4)^2 $: at x = -3, y = -1 (negative ✓), at x = -4, y = 0 (zero ✓), at x = -5, y = -1 (negative ✓).

Parabola Transformations with Vertex Translation

Which parabola is the translation of the graph of the function $y=-x^2$

and is negative in all domains except $x=-4$ ?

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Find the correct function according to the data

00:03 We'll use the negative domain, indicating that the parabola opens downward

00:08 Meaning a negative coefficient for X squared

00:12 Let's draw the function based on intersection points and parabola type

00:17 Left shift depends on term P

00:21 We'll substitute in the parabola formula and solve to find the function

00:26 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 parabola is the translation of the graph of the function $y=-x^2$

and is negative in all domains except $x=-4$ ?

Step-by-step solution

To solve this problem, consider the following steps:

Step 1: Recognize that translating a parabola horizontally involves modifying its vertex.
Step 2: The original function $y = -x^2$ has its vertex at $(0, 0)$ , and it is entirely non-positive.
Step 3: To make the parabola non-negative only at $x = -4$ , translate this parabola horizontally to have its vertex at $(-4, 0)$ .

Let us translate $y = -x^2$ by moving it 4 units to the left, resulting in:

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

This new equation $y = -(x + 4)^2$ translates the parabola left by 4 units, positioning the vertex at $(-4, 0)$ .

Since the coefficient of $(x + 4)^2$ is negative, the parabola is downward opening and is negative anywhere except at its vertex.

Therefore, the translated parabola is indeed negative throughout its domain except at $x = -4$ , which satisfies the problem's conditions.

Thus, the parabola is correctly represented by the expression $\mathbf{y = -(x + 4)^2}$ .

Final Answer

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

Key Points to Remember

Essential concepts to master this topic

Translation Rule: Replace x with (x + h) to shift left by h units
Technique: For vertex at (-4,0), use y = -(x + 4)² form
Check: Verify parabola equals zero only at x = -4 and negative elsewhere ✓

Common Mistakes

Avoid these frequent errors

Confusing direction of horizontal translation
Don't use y = -(x - 4)² to shift left = parabola shifts right instead! The sign inside parentheses is opposite to the direction. Always use y = -(x + 4)² to move the vertex 4 units left to x = -4.

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 it (x + 4) and not (x - 4) to move left?

This is counterintuitive but crucial! To shift left by 4 units, you replace x with (x + 4). Think of it this way: when x = -4, the expression (x + 4) equals zero, making the vertex at x = -4.

How do I know the parabola opens downward?

Look at the coefficient in front! In $y = -(x + 4)^2$ , the negative sign means it opens downward. Positive coefficient = opens up, negative coefficient = opens down.

What does 'negative everywhere except at x = -4' mean?

It means the parabola has y-values less than zero for all x-values except x = -4, where y = 0. At the vertex (-4, 0), the parabola touches the x-axis but never goes above it.

Could the answer be y = (x + 4)² instead?

No! That parabola opens upward and is positive everywhere except at x = -4. We need a downward-opening parabola that's negative everywhere except the vertex.

How can I verify my answer is correct?

Test a few points! For $y = -(x + 4)^2$ : at x = -3, y = -1 (negative ✓), at x = -4, y = 0 (zero ✓), at x = -5, y = -1 (negative ✓).

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