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

Find the ascending area of the function

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

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: $6 < x$.