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The graph of h(x) is a translation of f (x) = RootIndex 3 StartRoot x EndRoot. On a coordinate plane, a cube root function goes through (negative 3, negative 1), has an inflection point at (negative 2, 0), and goes through (negative 1, 1). Which equation represents h(x)?

The graph of h x is a translation of f x RootIndex 3 StartRoot x EndRoot On a coordinate plane a cube root function goes

Answer :

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).

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