Finding Anchor and Reference Point of a Cube Root Function.
By Sarah Smith
My brother sent me a question from his Pre-Calculus Study Guide. I have been studying math for $4$ years and don't ever recall seeing a question like this.
The question:
Graph $f(x) = (-1/2)(x+1)^{1/3} - 2$ on the axis provided. You must include the anchor and reference point to receive full credit.
Any advice on how to approach would be appreciated.
$\endgroup$ 32 Answers
$\begingroup$First plot $x^{1/3}$. It goes through the origin.
The change to $(x+1)^{1/3}$ moves it $1$ to the left so it goes through $(-1, 0)$.
The change to $-\frac12(x+1)^{1/3}$ turns it upside-down and compresses is vertically by a factor of $2$.
The final change to $-\frac12(x+1)^{1/3}-2$ moves it down by 2.
And that's the way it is.
$\endgroup$ 3 $\begingroup$In my son's textbook, it says for the general parent equation, $$f(x)=a\sqrt[3]{x-h} + k$$ the reference point is (1,1) which is equal to (1 + h, a + k). It's a quicker way maybe to solve for the inflection point $(h,k)$ and $a$.
So for equation $$f(x)=\sqrt[3](x)$$ the reference point is (1,1) and $h=0$, $k=0$ and $a=1$
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