Answer:
The correct option is;
[tex]h(x) = \sqrt[3]{x + 2}[/tex]
Step-by-step explanation:
Given that h(x) is a translation of f(x) = ∛x
From the points on the graph, given that the function goes through (-1, 1) and (-3, -1) we have;
When x = -1, h(x) = 1
When x = -3, h(x) = -1
h''(x) = (-2, 0)
Which gives
d²(∛(x + a))/dx²= [tex]-\left ( \dfrac{2}{9} \cdot \left (x + a \right )^{\dfrac{-5}{3}}\right )[/tex], have coordinates (-2, 0)
When h(x) = 0, x = -2 which gives;
[tex]-\left ( \dfrac{2}{9} \cdot \left (-2 + a \right )^{\dfrac{-5}{3}}\right ) = 0[/tex]
Therefore, a = (0/(-2/9))^(-3/5) + 2
a = 2
The translation is h(x) = [tex]\sqrt[3]{x + 2}[/tex]
We check, that when, x = -1, y = 1 which gives;
h(x) = [tex]\sqrt[3]{-1 + 2} = \sqrt[3]{1} = 1[/tex] which satisfies the condition that h(x) passes through the point (-1, 1)
For the point (-3, -1), we have;
h(x) = [tex]\sqrt[3]{-3 + 2} = \sqrt[3]{-1} = -1[/tex]
Therefore, the equation, h(x) = [tex]\sqrt[3]{x + 2}[/tex] passes through the points (-1, 1) and (-3, -1) and has an inflection point at (-2, 0).
