# How do you find the points where the graph of the function #(x-2)(x^2+1) / (x+3)# has horizontal tangents and what is the equation?

See explanation

graph{y(x+3)-(x^2+1)(x-2)=0 [-5, 5, -2.5, 2.5]} graph{y(x+3)-(1+x^2)(x-2)=0 [-320, 320, -160, 160]}

It is evident from the first graph that

there is a point of inflexion near x = 1.

The second contracted graph reveals the missing part on the left

that discloses a horizontal tangent that is close to y = 95.

this horizontal tangent.

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To find the points where the graph of the function has horizontal tangents, we need to find the values of x where the derivative of the function is equal to zero.

First, let's find the derivative of the function. Using the quotient rule, we have:

f'(x) = [(x+3)(2x) - (x-2)(x^2+1)] / (x+3)^2

Next, we set the derivative equal to zero and solve for x:

[(x+3)(2x) - (x-2)(x^2+1)] / (x+3)^2 = 0

Simplifying the equation, we get:

2x(x+3) - (x-2)(x^2+1) = 0

Expanding and rearranging terms, we have:

2x^2 + 6x - x^3 - 2x^2 + x + 2 = 0

Combining like terms, we get:

-x^3 + 7x + 2 = 0

To solve this equation, we can use numerical methods or graphing calculators. The solutions to this equation are approximately x ≈ -2.532, x ≈ 0.532, and x ≈ 3.532.

Therefore, the points where the graph of the function has horizontal tangents are approximately (-2.532, f(-2.532)), (0.532, f(0.532)), and (3.532, f(3.532)).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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