Question

choose the inverse of $y = x^2 - 10x$.
$\circ$ $y = \pm \sqrt{x - 25} - 5$
$\circ$ $y = \pm \sqrt{x - 25} + 5$
$\circ$ $y = \pm \sqrt{x + 25} - 5$
$\circ$ $y = \pm \sqrt{x + 25} + 5$
done

Explanation:

Step1: Swap x and y

To find the inverse, we first swap \( x \) and \( y \) in the equation \( y = x^2 - 10x \). So we get \( x = y^2 - 10y \).

Step2: Complete the square for y

We complete the square for the quadratic in \( y \). The quadratic is \( y^2 - 10y \). To complete the square, we take half of the coefficient of \( y \) (which is \( - 10\)), square it, and add and subtract it. Half of \( - 10\) is \( - 5\), and squaring it gives \( 25 \). So we rewrite the equation as:

$$ x=y^{2}-10y + 25-25 $$

The first three terms \( y^{2}-10y + 25\) is a perfect square trinomial, which factors to \( (y - 5)^2 \). So the equation becomes:

$$ x=(y - 5)^2-25 $$

Step3: Solve for y

First, we add \( 25 \) to both sides of the equation:

$$ x + 25=(y - 5)^2 $$

Then, we take the square root of both sides. Remember that when we take the square root, we have a positive and negative solution:

$$ \pm\sqrt{x + 25}=y - 5 $$

Finally, we add \( 5 \) to both sides to solve for \( y \):

$$ y=\pm\sqrt{x + 25}+5 $$

Answer:

\( y=\pm\sqrt{x + 25}+5 \) (the fourth option)