Find the Ascending Area of y=(x-6)² + 4: Quadratic Function Analysis

Q: How do I know which direction the parabola opens?

Look at the coefficient of the squared term! In $ y = (x-6)^2 + 4 $, the coefficient is +1 (positive), so the parabola opens upward like a U-shape.

Q: What does 'ascending area' mean?

Ascending area means the interval where the function is increasing - where y-values get larger as x-values get larger. It's the upward-sloping part of the graph.

Q: Why is the vertex at (6, 4) and not (-6, 4)?

In vertex form $ y = a(x-h)^2 + k $, the vertex is at (h, k). Since we have $ (x-6)^2 $, we get h = 6, not -6. The sign is opposite of what's inside the parentheses!

Q: How can I verify that x > 6 is the increasing interval?

Pick test points! Try x = 5 and x = 7:At x = 5: $ y = (5-6)^2 + 4 = 1 + 4 = 5 $At x = 7: $ y = (7-6)^2 + 4 = 1 + 4 = 5 $Wait, both equal 5? That's because they're equidistant from the vertex! Try x = 8: $ y = (8-6)^2 + 4 = 8 $ - now it's increasing!

Q: What if the coefficient was negative like y = -(x-6)² + 4?

Then the parabola would open downward and the function would be decreasing for x > 6 instead! The negative sign flips everything upside down.

Quadratic Functions with Vertex Form Analysis

Find the ascending area of the function

$y=(x-6)^2+4$

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Understand the problem

Find the ascending area of the function

$y=(x-6)^2+4$

Step-by-step solution

To solve this problem, we first note that the function is expressed in vertex form as $y = (x-6)^2 + 4$ . The vertex form of a quadratic function is $y = a(x - p)^2 + k$ , where the vertex is located at $(p, k)$ . For our function, this vertex is at $(6, 4)$ .

Since the coefficient of $(x-6)^2$ is positive, the parabola opens upwards. This means that the lowest point on the parabola is the vertex, making $x = 6$ the axis of symmetry and the point at which the function changes from decreasing to increasing.

Thus, the function is increasing on the interval where $x > 6$ . When $x$ is greater than 6, the value of $y$ increases as $x$ increases.

By examining the multiple-choice answers, we find that the choice that matches $6 < x$ is the correct one. Therefore, the ascending area of the function is for $x > 6$ .

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

Final Answer

$6 < x$

Key Points to Remember

Essential concepts to master this topic

Vertex Form: $y = a(x-h)^2 + k$ has vertex at (h,k)
Technique: When a > 0, parabola opens upward and increases for x > h
Check: Test x = 7: $y = (7-6)^2 + 4 = 5 > 4$ ✓

Common Mistakes

Avoid these frequent errors

Confusing the vertex x-coordinate with the inequality direction
Don't think the function increases for x < 6 just because you see (x-6) = the function increases for x > 6 because the parabola opens upward! The squared term makes the function decrease before the vertex and increase after it. Always identify whether a > 0 (opens up) or a < 0 (opens down) first.

Practice Quiz

Test your knowledge with interactive questions

Find the corresponding algebraic representation of the drawing:

FAQ

Everything you need to know about this question

How do I know which direction the parabola opens?

Look at the coefficient of the squared term! In $y = (x-6)^2 + 4$ , the coefficient is +1 (positive), so the parabola opens upward like a U-shape.

What does 'ascending area' mean?

Ascending area means the interval where the function is increasing - where y-values get larger as x-values get larger. It's the upward-sloping part of the graph.

Why is the vertex at (6, 4) and not (-6, 4)?

In vertex form $y = a(x-h)^2 + k$ , the vertex is at (h, k). Since we have $(x-6)^2$ , we get h = 6, not -6. The sign is opposite of what's inside the parentheses!

How can I verify that x > 6 is the increasing interval?

Pick test points! Try x = 5 and x = 7:

At x = 5: $y = (5-6)^2 + 4 = 1 + 4 = 5$
At x = 7: $y = (7-6)^2 + 4 = 1 + 4 = 5$

Wait, both equal 5? That's because they're equidistant from the vertex! Try x = 8: $y = (8-6)^2 + 4 = 8$ - now it's increasing!

What if the coefficient was negative like y = -(x-6)² + 4?

Then the parabola would open downward and the function would be decreasing for x > 6 instead! The negative sign flips everything upside down.

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