Question

graph the equation shown below by transforming the given graph of the parent function.
$y = (-x)^3$

Explanation:

Step1: Analyze the parent function

The parent function of a cubic function is \( y = x^3 \). Its graph is symmetric about the origin (odd function), increasing throughout the domain.

Step2: Analyze the transformation

For the function \( y = (-x)^3 \), we can simplify it: \( y = (-x)^3=-x^3 \) (since \((-a)^n=-a^n\) when \(n\) is odd). The transformation from \( y = x^3 \) to \( y=-x^3 \) is a reflection over the \(y\)-axis (or equivalently, a reflection over the \(x\)-axis for odd functions, but in this case, replacing \(x\) with \(-x\) gives a reflection over the \(y\)-axis).

Step3: Determine key points transformation

• For the parent function \( y = x^3 \), let's take some key points:
• When \(x = -2\), \(y=(-2)^3=-8\)
• When \(x=-1\), \(y = (-1)^3=-1\)
• When \(x = 0\), \(y=0^3 = 0\)
• When \(x = 1\), \(y=1^3=1\)
• When \(x = 2\), \(y=2^3 = 8\)
• For the function \( y=(-x)^3=-x^3 \), we substitute \(x\) with \(-x\) in terms of the input - output relationship. So if we have a point \((x,y)\) on \(y = x^3\), the corresponding point on \(y=(-x)^3\) will be \((-x,y)\) (since \(y = (-(-x))^3=x^3\) wait, no, let's recast. Let's compute \(y = (-x)^3\) for the same \(x\) values:
• When \(x=-2\), \(y=(-(-2))^3=2^3 = 8\)
• When \(x=-1\), \(y=(-(-1))^3=1^3 = 1\)
• When \(x = 0\), \(y=(-0)^3=0\)
• When \(x = 1\), \(y=(-1)^3=-1\)
• When \(x = 2\), \(y=(-2)^3=-8\)

So the graph of \(y = (-x)^3\) is the reflection of \(y = x^3\) over the \(y\)-axis. To graph it, we can take the points of the parent function \(y = x^3\) (like \((-2,-8)\), \((-1,-1)\), \((0,0)\), \((1,1)\), \((2,8)\)) and reflect them over the \(y\)-axis. The reflected points will be \((2,-8)\), \((1,-1)\), \((0,0)\), \((-1,1)\), \((-2,8)\). Then we plot these points and draw the curve.

Answer:

To graph \(y = (-x)^3\), reflect the graph of the parent function \(y=x^3\) over the \(y\)-axis. The key points transformation: for a point \((x,y)\) on \(y = x^3\), the corresponding point on \(y=(-x)^3\) is \((-x,y)\) (or by calculating \(y = (-x)^3\) directly for \(x\) values, we get points like \((-2,8)\), \((-1,1)\), \((0,0)\), \((1,-1)\), \((2,-8)\) which are the reflection of \((2,8)\), \((1,1)\), \((0,0)\), \((-1,-1)\), \((-2,-8)\) over the \(y\)-axis). Then plot these points and draw the cubic curve.