Find the Translation of y=-x² with Maximum Value at x=2

Q: How do I know the parabola opens downward?

Look at the coefficient of the squared term. In $ y = -(x-2)^2 $, we have a negative sign in front, so it opens downward just like the original $ y = -x^2 $.

Q: Could the answer be one of the upward-opening parabolas?

No! The options $ y = (x-2)^2 $ and $ y = (x+2)^2 $ open upward, so they're positive except at their vertex, not negative.

Q: How do I check my answer?

Test a few points! For $ y = -(x-2)^2 $: at x=1, y=-1 (negative ✓); at x=2, y=0 (zero, not negative ✓); at x=3, y=-1 (negative ✓).

Parabola Translations with Vertex Conditions

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

and is negative in all domains except $x=2$ ?

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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 Using the negative domain, we understand that the parabola is sad (opens downward):

00:08 Meaning a negative coefficient for X squared

00:11 Let's draw the function according to the intersection points and type of parabola

00:16 Right shift depends on term P

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

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

and is negative in all domains except $x=2$ ?

Step-by-step solution

The problem requires us to find a translated version of the parabola $y = -x^2$ such that the resulting function is negative for all $x$ except for a specific point, $x = 2$ . Here’s how we address this :

The function $y = -x^2$ is an upside-down parabola centered at the origin $(0, 0)$ .
We want the vertex to be at $x = 2$ . For a horizontal translation to the right by 2 units, we use $x - 2$ inside the function, leading to the new function $y = -(x - 2)^2$ .
This translation moves the vertex to the point $(2, 0)$ and ensures that the parabola will still open downwards, being negative for all $x \neq 2$ .

Thus, the parabola described by the function $y = -(x - 2)^2$ is what fulfills all the required conditions, including negativity in all domains except at $x = 2$ .

The correct choice, given the options, is $y = -(x - 2)^2$ , identified as Choice 3.

Therefore, the solution to the problem is $y = -(x - 2)^2$ .

Final Answer

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

Key Points to Remember

Essential concepts to master this topic

Translation Rule: Replace x with (x-h) to shift h units right
Technique: For vertex at x=2, use $y = -(x-2)^2$
Check: At x=2: $y = -(2-2)^2 = 0$ , elsewhere negative ✓

Common Mistakes

Avoid these frequent errors

Confusing horizontal translation direction
Don't think (x-2) moves left because of the minus sign = backwards translation! The minus is inside parentheses, so (x-2) actually shifts the graph 2 units RIGHT to put the vertex at x=2. Always remember: (x-h) moves h units in the positive direction.

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 does (x-2) move the parabola to the right instead of left?

This confuses many students! Think of it this way: the vertex occurs when the expression inside parentheses equals zero. For (x-2) to equal zero, x must equal 2. So the vertex moves to x=2, which is to the right.

How do I know the parabola opens downward?

Look at the coefficient of the squared term. In $y = -(x-2)^2$ , we have a negative sign in front, so it opens downward just like the original $y = -x^2$ .

What does 'negative in all domains except x=2' mean?

This means the y-values are negative (below the x-axis) everywhere except at x=2, where y=0. Since the parabola opens downward with vertex at (2,0), this condition is satisfied.

Could the answer be one of the upward-opening parabolas?

No! The options $y = (x-2)^2$ and $y = (x+2)^2$ open upward, so they're positive except at their vertex, not negative.

How do I check my answer?

Test a few points! For $y = -(x-2)^2$ : at x=1, y=-1 (negative ✓); at x=2, y=0 (zero, not negative ✓); at x=3, y=-1 (negative ✓).

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