Given the function y=(x^2-5/x+3), answer the following questions.

a) Find dy/dx. Simplify your answer
b) Find the coordinates of any points on the graph of y where the tangent line is horizontal. Show your work.
c) Are there any points on the graph of y where the tangent line is parallel to the line y=x+7? Explain your answers.

y' = (x^2+6x+5)/(x+3)^2 = (x+1)(x+5)/(x+3)^2

or, you could say y = x-3+4/(x+3), so
y' = 1 - 4/(x+3)^2

so, where is y'=0? x = -1 or -5

where is y'=1? nowhere, since 1-4/(x+3)^2 < 1 for all x. Note that y has a slant asymptote with slope=1.

a) To find dy/dx, we can use the quotient rule for differentiation. The quotient rule states that for a function u/v, the derivative is given by (v * du/dx - u * dv/dx) / v^2.

In this case, u = x^2 - 5 and v = x + 3. Let's differentiate u and v separately.

du/dx = d/dx(x^2 - 5)
= 2x

dv/dx = d/dx(x + 3)
= 1

Now, substituting these values into the quotient rule formula, we get:

dy/dx = (v * du/dx - u * dv/dx) / v^2
= ((x + 3) * (2x) - (x^2 - 5) * 1) / (x + 3)^2
= (2x(x + 3) - (x^2 - 5))/(x + 3)^2
= (2x^2 + 6x - x^2 + 5)/(x + 3)^2
= (x^2 + 6x + 5)/(x + 3)^2

Therefore, dy/dx = (x^2 + 6x + 5)/(x + 3)^2.

b) To find the coordinates of any points on the graph of y where the tangent line is horizontal, we need to find the values of x that make dy/dx equal to zero.

So, setting the numerator equal to zero:

x^2 + 6x + 5 = 0

We can solve this quadratic equation by factoring or using the quadratic formula. Factoring is not possible in this case, so let's use the quadratic formula:

x = (-6 ± √(6^2 - 4(1)(5))) / (2 * 1)
x = (-6 ± √(36 - 20)) / 2
x = (-6 ± √16) / 2
x = (-6 ± 4) / 2

Simplifying further, we get two possible values for x:

x₁ = (-6 + 4) / 2 = -1
x₂ = (-6 - 4) / 2 = -5

Now, substitute these values of x back into the original function y = (x^2 - 5) / (x + 3) to find the corresponding y-coordinates:

For x = -1:
y = (-1^2 - 5) / (-1 + 3) = (-1 - 5) / 2 = -3

For x = -5:
y = (-5^2 - 5) / (-5 + 3) = (25 - 5) / (-2) = -10

Therefore, the coordinates of the points where the tangent line is horizontal are (-1, -3) and (-5, -10).

c) To determine if there are any points on the graph of y where the tangent line is parallel to the line y = x + 7, we need to check if the slopes of y and the line y = x + 7 are equal.

The slope of the tangent line at any point can be found by evaluating dy/dx at that point.

The slope of the line y = x + 7 is 1.

So, we need to find points on the graph of y where dy/dx equals 1.

Setting dy/dx equal to 1:

(x^2 + 6x + 5)/(x + 3)^2 = 1

Multiplying both sides by (x + 3)^2:

x^2 + 6x + 5 = (x + 3)^2

Expanding the right side of the equation:

x^2 + 6x + 5 = x^2 + 6x + 9

Simplifying:

x^2 + 6x + 5 - x^2 - 6x - 9 = 0

-4 = 0

However, this equation has no solution. Therefore, there are no points on the graph of y where the tangent line is parallel to the line y = x + 7.