Find the Area Above y=(x-6)²+3: Shifted Quadratic Function

Q: How do I find the vertex from vertex form?

In vertex form $ y = (x-h)^2 + k $, the vertex is simply (h, k). For $ y = (x-6)^2 + 3 $, the vertex is (6, 3).

Q: Why does the parabola increase after x = 6?

Since the coefficient of $ (x-6)^2 $ is positive (it's 1), the parabola opens upward. This means it decreases before the vertex and increases after the vertex at x = 6.

Q: What does 'ascending area' mean?

Ascending area means the region where the function is increasing - where y-values get larger as x increases. For this parabola, that's when $ x > 6 $.

Q: How can I tell if a parabola opens up or down?

Look at the coefficient of the squared term! If it's positive, the parabola opens upward. If it's negative, it opens downward. Here, the coefficient is +1, so it opens up.

Q: What if the question asked for the descending area?

The descending area would be where the function decreases. For this upward-opening parabola, that would be \( x - to the left of the vertex.

Quadratic Functions with Vertex Identification

Find the ascending area of the function

$y=(x-6)^2+3$

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Understand the problem

Find the ascending area of the function

$y=(x-6)^2+3$

Step-by-step solution

The function given is $y = (x-6)^2 + 3$ . This is a standard parabola with vertex form $(x - p)^2 + k$ .

Step-by-step solution:

Identify the vertex: The vertex form of a parabola is given as $y = (x-p)^2 + k$ , where $(p, k)$ is the vertex. For this function, $p = 6$ and $k = 3$ , so the vertex is $(6, 3)$ .
Determine the direction of the parabola: The parabola opens upwards because the coefficient of $(x-6)^2$ is positive (specifically, it equals 1). This means the parabola decreases on the left of the vertex and increases on the right.
Determine the domain for increasing values: Since the vertex is at $x = 6$ and the parabola opens upwards, the graph of the function is increasing for all $x$ values greater than 6. Thus, for $x > 6$ , the function $y = (x-6)^2 + 3$ is increasing.

Therefore, the solution to the problem is $x > 6$ .

Thus, the correct answer from the given choices is: $6 < x$ .

Final Answer

$6 < x$

Key Points to Remember

Essential concepts to master this topic

Vertex Form: In $y = (x-h)^2 + k$ , vertex is (h, k)
Technique: From $y = (x-6)^2 + 3$ , vertex is (6, 3)
Check: Positive coefficient means increasing for x > 6 ✓

Common Mistakes

Avoid these frequent errors

Confusing vertex coordinates with increasing intervals
Don't assume the vertex x-value tells you where the function increases = wrong intervals! The vertex only shows the turning point. Always check: upward parabolas increase to the right of the vertex, so x > 6 means ascending area.

Practice Quiz

Test your knowledge with interactive questions

Choose the equation that represents the function

$$ y=-x^2 $$

moved 3 spaces to the left

and 4 spaces up.

FAQ

Everything you need to know about this question

How do I find the vertex from vertex form?

In vertex form $y = (x-h)^2 + k$ , the vertex is simply (h, k). For $y = (x-6)^2 + 3$ , the vertex is (6, 3).

Why does the parabola increase after x = 6?

Since the coefficient of $(x-6)^2$ is positive (it's 1), the parabola opens upward. This means it decreases before the vertex and increases after the vertex at x = 6.

What does 'ascending area' mean?

Ascending area means the region where the function is increasing - where y-values get larger as x increases. For this parabola, that's when $x > 6$ .

How can I tell if a parabola opens up or down?

Look at the coefficient of the squared term! If it's positive, the parabola opens upward. If it's negative, it opens downward. Here, the coefficient is +1, so it opens up.

What if the question asked for the descending area?

The descending area would be where the function decreases. For this upward-opening parabola, that would be $x < 6$ - to the left of the vertex.

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