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

Video Solution

Solution Steps

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 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 $\mathbf{y = -(x + 4)^2}$ .