Question

write the expression to complete the equation for the function represented by the graph shown below.

(graph description: a coordinate plane with x-axis from -6 to 6 and y-axis from -6 to 6. a pink graph is shown, and there is a box at the bottom with y = and a blank to fill.)

Explanation:

Step1: Identify the vertex and slope

The graph is a V - shaped graph, so it is an absolute - value function. The vertex of the absolute - value graph is at \((3, - 3)\)? Wait, no, let's look at the graph again. Wait, the pink graph: let's find two points. Let's see, when \(x = 0\), \(y=-6\)? Wait, no, the graph: let's find the vertex. The vertex is at \((3, - 3)\)? Wait, no, let's check the points. Wait, the left side: from \((0, - 6)\) to \((3, - 3)\), the slope is \(\frac{-3 - (-6)}{3-0}=\frac{3}{3} = 1\). The right side: from \((3, - 3)\) to \((6, - 6)\), the slope is \(\frac{-6-(-3)}{6 - 3}=\frac{-3}{3}=-1\). So the absolute - value function has the form \(y=-|x - 3|-3\)? Wait, no, wait, let's check the vertex. Wait, maybe I made a mistake. Wait, the standard form of an absolute - value function is \(y=a|x - h|+k\), where \((h,k)\) is the vertex. Let's find the vertex. Looking at the graph, the peak (vertex) is at \((3, - 3)\)? Wait, no, when \(x = 3\), \(y=-3\), and when \(x = 0\), \(y=-6\), when \(x = 6\), \(y=-6\). Wait, no, let's recalculate. Wait, the left line: from \((0, - 6)\) to \((3, - 3)\): slope \(m=\frac{-3+6}{3 - 0}=\frac{3}{3}=1\). The right line: from \((3, - 3)\) to \((6, - 6)\): slope \(m=\frac{-6 + 3}{6-3}=\frac{-3}{3}=-1\). So the function is \(y=-|x - 3|-3\)? Wait, no, let's plug in \(x = 3\): \(y=-|0|-3=-3\), correct. \(x = 0\): \(y=-| - 3|-3=-3 - 3=-6\), correct. \(x = 6\): \(y=-|3|-3=-3 - 3=-6\), correct. Wait, but maybe the vertex is at \((3, - 3)\). Wait, but let's check again. Alternatively, maybe the function is \(y=-|x|-6\)? No, when \(x = 3\), \(y=-|3|-6=-9\), which is not correct. Wait, I think I made a mistake in identifying the points. Wait, the graph: let's look at the grid. Each square is 1 unit. The pink graph: when \(x = 0\), \(y=-6\); when \(x = 3\), \(y=-3\); when \(x = 6\), \(y=-6\). So the vertex is at \((3, - 3)\). So the equation is \(y=-|x - 3|-3\)? Wait, no, the standard form is \(y=a|x - h|+k\). Here, \(h = 3\), \(k=-3\), and \(a=-1\) (because the graph opens downward). So \(y=-|x - 3|-3\)? Wait, but when \(x = 0\), \(y=-| - 3|-3=-3 - 3=-6\), which matches. When \(x = 3\), \(y=-0 - 3=-3\), which matches. When \(x = 6\), \(y=-|3|-3=-3 - 3=-6\), which matches. Alternatively, maybe the function is \(y=-|x|-6\)? No, when \(x = 3\), \(y=-3 - 6=-9 eq - 3\). So the correct function is \(y=-|x - 3|-3\)? Wait, no, wait, maybe the vertex is at \((0, - 6)\)? No, the graph is symmetric about \(x = 3\). Wait, the distance from \(x = 0\) to \(x = 6\) is 6, so the axis of symmetry is \(x = 3\). So the vertex is at \((3, - 3)\). So the equation is \(y=-|x - 3|-3\). Wait, but let's check the slope again. From \(x = 0\) to \(x = 3\), \(y\) goes from \(-6\) to \(-3\), so the slope is \(\frac{-3+6}{3-0}=1\). From \(x = 3\) to \(x = 6\), \(y\) goes from \(-3\) to \(-6\), slope is \(\frac{-6 + 3}{6-3}=-1\). So the function is \(y=-|x - 3|-3\). Wait, but maybe I made a mistake. Alternatively, the function could be \(y=-|x|-6\) if the vertex is at \((0, - 6)\), but that doesn't match the symmetry. Wait, no, the graph is symmetric about \(x = 3\), so the vertex is at \((3, - 3)\). So the equation is \(y=-|x - 3|-3\). Wait, but let's check another point. When \(x = 3\), \(y=-3\), correct. When \(x = 1\), \(y=-|1 - 3|-3=-| - 2|-3=-2 - 3=-5\). Let's see if that's on the graph. From \(x = 0\) (\(y=-6\)) to \(x = 3\) (\(y=-3\)), at \(x = 1\), \(y=-6+(1 - 0)\times1=-5\), which is correct. At \(x = 2\), \(y=-6+(2 - 0)\times1=-4\), and using the formula \(y=-|2 - 3|-3=-| - 1|-3=-1 - 3=-4\), correct. So the equation is \(y=-|x - 3|-3\)? Wait, no, wait, maybe the vertex is at \((3, - 3)\), so the equation is \(y=-|x - 3|-3\).

Wait, another approach: the graph is a reflection of the standard \(y = |x|\) graph, shifted and scaled. The standard \(y=|x|\) has vertex at \((0,0)\), opens up. This graph opens down, has vertex at \((3, - 3)\). So the equation is \(y=-|x - 3|-3\).

Step2: Verify the equation

Let's check \(x = 0\): \(y=-|0 - 3|-3=-3 - 3=-6\), which matches the graph (when \(x = 0\), \(y=-6\)). \(x = 3\): \(y=-|0|-3=-3\), which is the vertex. \(x = 6\): \(y=-|6 - 3|-3=-3 - 3=-6\), which matches the graph (when \(x = 6\), \(y=-6\)). \(x = 1\): \(y=-|1 - 3|-3=-2 - 3=-5\), and from the left line (slope 1 from \((0, - 6)\) to \((3, - 3)\)), at \(x = 1\), \(y=-6+(1)\times1=-5\), correct. \(x = 2\): \(y=-|2 - 3|-3=-1 - 3=-4\), and from the left line, \(y=-6+(2)\times1=-4\), correct. \(x = 4\): \(y=-|4 - 3|-3=-1 - 3=-4\), and from the right line (slope - 1 from \((3, - 3)\) to \((6, - 6)\)), at \(x = 4\), \(y=-3+(4 - 3)\times(-1)=-3 - 1=-4\), correct. \(x = 5\): \(y=-|5 - 3|-3=-2 - 3=-5\), and from the right line, \(y=-3+(5 - 3)\times(-1)=-3 - 2=-5\), correct. So the equation is \(y=-|x - 3|-3\).

Wait, but maybe I made a mistake in the vertex. Wait, let's look at the graph again. The pink graph: when \(x = 0\), \(y=-6\); when \(x = 3\), \(y=-3\); when \(x = 6\), \(y=-6\). So it's symmetric about \(x = 3\), vertex at \((3, - 3)\), and the slope of the left side is 1, right side is - 1, so the equation is \(y=-|x - 3|-3\).

Answer:

\(y=-|x - 3|-3\)