Transformations Of Quadratic Function $y= X^2-6x+10$

Alright, let's dive into how the quadratic function $y = x^2 - 6x + 10$ transforms under different operations. We're going to explore reflections, translations, and rotations to see how the original function changes. So, buckle up, and let's get started!

a) Reflection About the X-Axis, Then Vertical Translation 3 Units Down

First, we'll tackle the reflection about the x-axis. When a function $y = f(x)$ is reflected about the x-axis, the new function becomes $y = -f(x)$. So, for our function $y = x^2 - 6x + 10$, the reflection about the x-axis gives us:

$y = -(x^2 - 6x + 10) = -x^2 + 6x - 10$

Now, we need to perform a vertical translation of 3 units down. This means we subtract 3 from the entire function:

$y = -x^2 + 6x - 10 - 3 = -x^2 + 6x - 13$

So, the final transformed function after reflection about the x-axis and a vertical translation of 3 units down is:

$y = -x^2 + 6x - 13$

In summary, reflecting the function $y = x^2 - 6x + 10$ across the x-axis inverts the sign of the entire function, resulting in $y = -x^2 + 6x - 10$. Subsequently, shifting this reflected function 3 units downward involves subtracting 3 from the function, leading to the final transformed function $y = -x^2 + 6x - 13$. This transformation effectively flips the parabola over the x-axis and moves it lower on the graph. Understanding these transformations is crucial for visualizing how quadratic functions behave under different operations, providing insights into their graphical representations and algebraic properties. The process involves two key steps: first, applying the reflection by negating the function, and second, applying the translation by adjusting the constant term. These techniques are fundamental in coordinate geometry and are widely used in various mathematical and engineering applications.

b) Reflection About the Line $y=1$, Followed by Reflection About the Line $x=3$

Now, let's handle the reflections about the lines $y = 1$ and $x = 3$.

Reflection About the Line $y = 1$

When reflecting about the line $y = 1$, we use the transformation $y' = 2(1) - y$. So, for our function $y = x^2 - 6x + 10$, the new $y'$ is:

$y' = 2(1) - (x^2 - 6x + 10) = 2 - x^2 + 6x - 10 = -x^2 + 6x - 8$

Reflection About the Line $x = 3$

Next, we reflect the transformed function $y' = -x^2 + 6x - 8$ about the line $x = 3$. This means we replace $x$ with $2(3) - x = 6 - x$:

$y'' = -(6 - x)^2 + 6(6 - x) - 8$

Expanding and simplifying:

$y'' = -(36 - 12x + x^2) + 36 - 6x - 8$ $y'' = -36 + 12x - x^2 + 36 - 6x - 8$ $y'' = -x^2 + 6x - 8$

So, the final transformed function after both reflections is:

$y = -x^2 + 6x - 8$

In summary, reflecting the original function $y = x^2 - 6x + 10$ about the line $y = 1$ involves replacing $y$ with $2 - y$, resulting in the intermediate function $y' = -x^2 + 6x - 8$. Subsequently, reflecting this intermediate function about the line $x = 3$ involves replacing $x$ with $6 - x$, leading to the final transformed function $y'' = -x^2 + 6x - 8$. This double reflection process effectively mirrors the parabola first across the horizontal line $y = 1$ and then across the vertical line $x = 3$. The key to these transformations lies in understanding how coordinates change with respect to the reflection lines. For the $y = 1$ reflection, the new y-coordinate is determined by subtracting the original y-coordinate from 2. For the $x = 3$ reflection, the new x-coordinate is determined by subtracting the original x-coordinate from 6. These techniques are crucial for geometric transformations and are used in various fields such as computer graphics and image processing. Understanding these principles allows for precise manipulation and analysis of functions in coordinate space.

c) Rotation $90^\*circ$ (Clockwise)

To rotate the function $y = x^2 - 6x + 10$ by $90^\*circ$ clockwise, we need to swap $x$ and $y$ and adjust the signs accordingly. A $90^\*circ$ clockwise rotation transforms $(x, y)$ to $(y, -x)$. So, we replace $x$ with $y$ and $y$ with $-x$ in the original equation:

$-x = y^2 - 6y + 10$

Now, we solve for $y$ to express it as a function of $x$:

$y^2 - 6y + (10 + x) = 0$

We can use the quadratic formula to solve for $y$:

$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In this case, $a = 1$, $b = -6$, and $c = 10 + x$. Plugging these values into the quadratic formula, we get:

$y = \frac{6 \pm \sqrt{(-6)^2 - 4(1)(10 + x)}}{2(1)}$ $y = \frac{6 \pm \sqrt{36 - 40 - 4x}}{2}$ $y = \frac{6 \pm \sqrt{-4 - 4x}}{2}$ $y = \frac{6 \pm 2\sqrt{-1 - x}}{2}$ $y = 3 \pm \sqrt{-1 - x}$

So, the rotated function is:

$y = 3 \pm \sqrt{-1 - x}$

In summary, rotating the function $y = x^2 - 6x + 10$ by $90^\*circ$ clockwise involves swapping the x and y variables and adjusting the signs according to the rotation transformation. Specifically, the transformation $(x, y) \rightarrow (y, -x)$ is applied, leading to the equation $-x = y^2 - 6y + 10$. To express $y$ as a function of $x$, we rearrange the equation into a quadratic form $y^2 - 6y + (10 + x) = 0$ and then apply the quadratic formula. The quadratic formula yields two possible solutions for $y$, given by $y = \frac{6 \pm \sqrt{(-6)^2 - 4(1)(10 + x)}}{2}$. Simplifying this expression leads to $y = 3 \pm \sqrt{-1 - x}$. Thus, the rotated function is represented by $y = 3 \pm \sqrt{-1 - x}$, indicating two possible branches due to the square root. This rotation significantly alters the original parabolic function, transforming it into a sideways-opening function with a restricted domain due to the negative value under the square root. Understanding such transformations is crucial in various fields such as physics, engineering, and computer graphics, where rotations are fundamental operations.

I hope this explanation helps you understand how the function $y = x^2 - 6x + 10$ transforms under these different operations! Let me know if you have any more questions.