Question

1. write the equation for the absolute value function that has a vertex at (4, -1) and the points (6, -2) and (-6, 2)

Explanation:

Step1: Recall vertex form

The vertex form of an absolute value function is $y = a|x - h| + k$, where $(h,k)$ is the vertex. Here, $(h,k)=(4,-1)$, so substitute into the form:
$y = a|x - 4| - 1$

Step2: Solve for $a$ using $(6,-2)$

Substitute $x=6$, $y=-2$ into the equation:
$-2 = a|6 - 4| - 1$
$-2 = 2a - 1$
$2a = -2 + 1 = -1$
$a = -\frac{1}{2}$

Step3: Verify with $(-6,2)$

Substitute $x=-6$, $a=-\frac{1}{2}$ into $y = a|x - 4| - 1$:
$y = -\frac{1}{2}|-6 - 4| - 1 = -\frac{1}{2}(10) - 1 = -5 - 1 = -6$
This does not match the given point $( -6,2)$, so use $( -6,2)$ to solve for $a$ instead:
$2 = a|-6 - 4| - 1$
$2 = 10a - 1$
$10a = 3$
$a = \frac{3}{10}$
Verify with $(6,-2)$:
$y = \frac{3}{10}|6 - 4| - 1 = \frac{3}{10}(2) - 1 = \frac{3}{5} - 1 = -\frac{2}{5}$, which does not match. Since the problem states the function passes through both points, we use the vertex and solve for the correct $a$ that satisfies both.
Set up system of equations:

1. $-2 = a|6 - 4| - 1 \implies 2a = -1 \implies a=-\frac{1}{2}$
2. $2 = a|-6 - 4| - 1 \implies 10a = 3 \implies a=\frac{3}{10}$

Since no single $a$ satisfies both, the function must be a piecewise absolute value function, but the standard form uses the vertex and one point. Assuming the problem intends the first valid pair, we use $(4,-1)$ and $(6,-2)$.

Answer:

$y = -\frac{1}{2}|x - 4| - 1$

Note: There is a contradiction with the point $(-6,2)$ not fitting the function derived from the vertex and $(6,-2)$. If forced to use all three, the function is not a standard absolute value function, but the most consistent standard form using the vertex and one given point is above.