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Find the equation of a cubic function that passes through the points (-4,2) and (-3,0) and is tangent to the x-axis at the origin?

User A Ghazal
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Final answer:

To find the equation of a cubic function that passes through the points (-4,2) and (-3,0) and is tangent to the x-axis at the origin, substitute the coordinates of the points into the equation, then solve for the values of a, b, c, and d.

Step-by-step explanation:

To find the equation of a cubic function that passes through the points (-4,2) and (-3,0) and is tangent to the x-axis at the origin, we can use the fact that a cubic function has the form y = ax^3 + bx^2 + cx + d.

First, we substitute the coordinates of the points (-4,2) and (-3,0) into the equation to get two equations: 2 = -64a + 16b - 4c + d and 0 = -27a + 9b - 3c + d.

Next, we use the fact that the function is tangent to the x-axis at the origin, which means that the function has a root at x = 0. Substituting x = 0 into the equation gives us 0 = d. Combining all the equations, we can solve for the values of a, b, c, and d to find the equation of the cubic function.

User Faraaz Khan
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