Question

svlc algebra 1a - standard (15260)
representing relationships

$x$	$f(x)$
$-4$	$2$
$-2$	$0$
$0$	$-2$
$2$	$-1$
$4$	$0$
$6$	$4$

which is a possible turning point for the continuous function $f(x)$?
options:
$(0, -2)$
$(4, 0)$
$(2, -1)$
$(-2, 0)$

Explanation:

Step1: Understand Turning Point

A turning point of a function is a point where the function changes its direction (from increasing to decreasing or vice - versa). To identify a possible turning point, we analyze the behavior of the function (increasing or decreasing) around each given point by looking at the values of \(x\) and \(f(x)\) in the table.

Step2: Analyze Behavior Around (0, - 2)

- For \(x=-2\), \(f(-2) = 0\); for \(x = 0\), \(f(0)=-2\); for \(x = 2\), \(f(2)=-1\).
- From \(x=-2\) to \(x = 0\) (moving from left to right, \(x\) increases), \(f(x)\) changes from \(0\) to \(-2\) (decreasing). From \(x = 0\) to \(x = 2\), \(f(x)\) changes from \(-2\) to \(-1\) (increasing). So the function changes from decreasing to increasing at \(x = 0\)? Wait, no, let's check other points.

Step3: Analyze Behavior Around (4, 0)

- For \(x = 2\), \(f(2)=-1\); for \(x = 4\), \(f(4)=0\); for \(x = 6\), \(f(6)=4\).
- From \(x = 2\) to \(x = 4\), \(f(x)\) changes from \(-1\) to \(0\) (increasing). From \(x = 4\) to \(x = 6\), \(f(x)\) changes from \(0\) to \(4\) (increasing). So the function is increasing on both sides of \(x = 4\), not a turning point.

Step4: Analyze Behavior Around (2, - 1)

- For \(x = 0\), \(f(0)=-2\); for \(x = 2\), \(f(2)=-1\); for \(x = 4\), \(f(4)=0\).
- From \(x = 0\) to \(x = 2\), \(f(x)\) changes from \(-2\) to \(-1\) (increasing). From \(x = 2\) to \(x = 4\), \(f(x)\) changes from \(-1\) to \(0\) (increasing). So the function is increasing on both sides of \(x = 2\), not a turning point.

Step5: Analyze Behavior Around (- 2, 0)

- For \(x=-4\), \(f(-4) = 2\); for \(x=-2\), \(f(-2)=0\); for \(x = 0\), \(f(0)=-2\).
- From \(x=-4\) to \(x=-2\) (increasing \(x\)), \(f(x)\) changes from \(2\) to \(0\) (decreasing). From \(x=-2\) to \(x = 0\) (increasing \(x\)), \(f(x)\) changes from \(0\) to \(-2\) (decreasing). Wait, no, let's re - evaluate the behavior around (2, - 1) again. Wait, maybe we made a mistake. Let's check the slope (rate of change) between points.

The slope between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(m=\frac{y_2 - y_1}{x_2 - x_1}\)

- Between \((-2,0)\) and \((0,-2)\): \(m=\frac{-2 - 0}{0-(-2)}=\frac{-2}{2}=-1\) (decreasing)
- Between \((0,-2)\) and \((2,-1)\): \(m=\frac{-1-(-2)}{2 - 0}=\frac{1}{2}\) (increasing)
- Between \((2,-1)\) and \((4,0)\): \(m=\frac{0 - (-1)}{4 - 2}=\frac{1}{2}\) (increasing)
- Between \((4,0)\) and \((6,4)\): \(m=\frac{4 - 0}{6 - 4}=\frac{4}{2}=2\) (increasing)
- Between \((-4,2)\) and \((-2,0)\): \(m=\frac{0 - 2}{-2-(-4)}=\frac{-2}{2}=-1\) (decreasing)
- Between \((-6,8)\) and \((-4,2)\): \(m=\frac{2 - 8}{-4-(-6)}=\frac{-6}{2}=-3\) (decreasing)

Wait, the function changes from decreasing (from \(x=-6\) to \(x = 0\) with slopes negative) to increasing (from \(x = 0\) to \(x = 6\) with slopes positive)? No, from \(x = 0\) to \(x = 2\) slope is \(1/2\), \(x = 2\) to \(x = 4\) slope is \(1/2\), \(x = 4\) to \(x = 6\) slope is \(2\). But around \((2,-1)\): before \((2,-1)\) (from \(0,-2\) to \(2,-1\)) the slope is \(1/2\) (increasing), after \((2,-1)\) (from \(2,-1\) to \(4,0\)) the slope is \(1/2\) (still increasing). Around \((4,0)\): before (from \(2,-1\) to \(4,0\)) slope \(1/2\) (increasing), after (from \(4,0\) to \(6,4\)) slope \(2\) (increasing). Around \((-2,0)\): before (from \(-4,2\) to \(-2,0\)) slope \(-1\) (decreasing), after (from \(-2,0\) to \(0,-2\)) slope \(-1\) (decreasing). Around \((0,-2)\): before (from \(-2,0\) to \(0,-2\)) slope \(-1\) (decreasing), after (from \(0,-2\) to \(2,-1\)) slope \(1/2\) (increasing). Ah! So at \(x = 0\), the function changes from decreasing (slope - 1) to increasing (slope \(1/2\)). Wait, but \((0,-2)\) is one of the options. Wait, no, let's check the options again. Wait, maybe we misread the options. The options are \((0,-2)\), \((4,0)\), \((2,-1)\), \((-2,0)\).

Wait, a turning point is where the function changes from increasing to decreasing or decreasing to increasing. Let's check the behavior of \(f(x)\) around each point:

- For \((0,-2)\):
- Left of \(x = 0\) (e.g., \(x=-2\), \(f(-2)=0\); \(x = 0\), \(f(0)=-2\)): as \(x\) increases from \(-2\) to \(0\), \(f(x)\) decreases (from \(0\) to \(-2\)).
- Right of \(x = 0\) (e.g., \(x = 0\), \(f(0)=-2\); \(x = 2\), \(f(2)=-1\)): as \(x\) increases from \(0\) to \(2\), \(f(x)\) increases (from \(-2\) to \(-1\)). So the function changes from decreasing to increasing at \(x = 0\), so \((0,-2)\) is a point where the function changes direction? Wait, no, maybe \((2,-1)\). Wait, no, between \((0,-2)\) and \((2,-1)\), the slope is positive (increasing), between \((2,-1)\) and \((4,0)\), the slope is positive (increasing). Between \((-2,0)\) and \((0,-2)\), slope is negative (decreasing), between \((0,-2)\) and \((2,-1)\), slope is positive (increasing). So at \(x = 0\), the function changes from decreasing to increasing. But wait, let's check the other points.

Wait, maybe the answer is \((2,-1)\)? No, let's check the values again. Wait, the function values:

- When \(x\) goes from \(-6\) to \(-4\): \(f(x)\) goes from \(8\) to \(2\) (decreasing)
- \(-4\) to \(-2\): \(2\) to \(0\) (decreasing)
- \(-2\) to \(0\): \(0\) to \(-2\) (decreasing)
- \(0\) to \(2\): \(-2\) to \(-1\) (increasing)
- \(2\) to \(4\): \(-1\) to \(0\) (increasing)
- \(4\) to \(6\): \(0\) to \(4\) (increasing)

Wait, so the function is decreasing from \(x=-\infty\) (in the domain we have) up to \(x = 0\), and then increasing from \(x = 0\) onwards? No, from \(0\) to \(2\) it's increasing, \(2\) to \(4\) increasing, \(4\) to \(6\) increasing. Wait, but from \(-6\) to \(0\) it's decreasing. So the point where it changes from decreasing to increasing is at \(x = 0\), so \((0,-2)\) is a turning point? But wait, let's check the options again. Wait, maybe I made a mistake. Let's check the slope between \((-2,0)\) and \((0,-2)\): slope is \(\frac{-2 - 0}{0 - (-2)}=-1\) (decreasing). Between \((0,-2)\) and \((2,-1)\): slope is \(\frac{-1-(-2)}{2 - 0}=\frac{1}{2}\) (increasing). So at \(x = 0\), the function changes from decreasing to increasing. So \((0,-2)\) is a turning point? But wait, the other option \((2,-1)\): between \((0,-2)\) and \((2,-1)\) slope is \(1/2\) (increasing), between \((2,-1)\) and \((4,0)\) slope is \(1/2\) (increasing). So no change. \((4,0)\): between \((2,-1)\) and \((4,0)\) slope \(1/2\) (increasing), between \((4,0)\) and \((6,4)\) slope \(2\) (increasing). \((-2,0)\): between \((-4,2)\) and \((-2,0)\) slope \(-1\) (decreasing), between \((-2,0)\) and \((0,-2)\) slope \(-1\) (decreasing). So the only point where the function changes from decreasing to increasing is at \((0,-2)\)? Wait, but maybe I messed up. Wait, let's check the value of \(f(x)\) around \(x = 2\). At \(x = 2\), \(f(x)=-1\). Before \(x = 2\) (e.g., \(x = 0\), \(f(0)=-2\)) and after \(x = 2\) (e.g., \(x = 4\), \(f(4)=0\)). So from \(x = 0\) to \(x = 2\), \(f(x)\) increases from \(-2\) to \(-1\), and from \(x = 2\) to \(x = 4\), \(f(x)\) increases from \(-1\) to \(0\). So no change. At \(x = 0\), before \(x = 0\) ( \(x=-2\), \(f(-2)=0\)) \(f(x)\) is decreasing (from \(0\) to \(-2\) as \(x\) goes from \(-2\) to \(0\)), and after \(x = 0\) ( \(x = 2\), \(f(2)=-1\)) \(f(x)\) is increasing (from \(-2\) to \(-1\) as \(x\) goes from \(0\) to \(2\)). So \((0,-2)\) is a turning point. Wait, but let's check the answer options again. Wait, maybe the correct answer is \((2,-1)\)? No, I think I made a mistake. Wait, let's recall that a turning point is a point where the derivative (slope) changes sign. So we can look at the slopes between consecutive points:

- From \((-6,8)\) to \((-4,2)\): slope \(m_1=\frac{2 - 8}{-4+6}=\frac{-6}{2}=-3\) (negative)
- From \((-4,2)\) to \((-2,0)\): slope \(m_2=\frac{0 - 2}{-2 + 4}=\frac{-2}{2}=-1\) (negative)
- From \((-2,0)\) to \((0,-2)\): slope \(m_3=\frac{-2-0}{0 + 2}=\frac{-2}{2}=-1\) (negative)
- From \((0,-2)\) to \((2,-1)\): slope \(m_4=\frac{-1 + 2}{2-0}=\frac{1}{2}\) (positive)
- From \((2,-1)\) to \((4,0)\): slope \(m_5=\frac{0 + 1}{4 - 2}=\frac{1}{2}\) (positive)
- From \((4,0)\) to \((6,4)\): slope \(m_6=\frac{4-0}{6 - 4}=\frac{4}{2}=2\) (positive)

So the slope changes from negative to positive at \(x = 0\) (between \(x=-2\) to \(x = 0\) slope is - 1 (negative) and between \(x = 0\) to \(x = 2\) slope is \(1/2\) (positive)). So the point \((0,-2)\) is a point where the slope changes sign, so it's a turning point. Wait, but let's check the options again. The options are \((0,-2)\), \((4,0)\), \((2,-1)\), \((-2,0)\). So \((0,-2)\) is the point where the function changes from decreasing to increasing, so it's a turning point.

Answer:

\((0,-2)\)