Verify if correct or not please =)

What is the slope of the normal to y= (x^2-1)/(x+5) at (1,0)

a) -3
b) -1/2
c) -1/3
d) 1/3

Answer is A, -3.

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What is the y-intercept of the line that is tangent to f(x)=4/(3x+1) at (-1, -2)?

a)-5
b) -3/4
c) 3/4
d) -2

Answer is -5?

Both are correct. Good job.

Thanks!

To verify the slope of the normal to the equation y = (x^2-1)/(x+5) at the point (1,0), we need to find the derivative of the function using the quotient rule.

1. Differentiate the numerator:

dy/dx = (2x)(x + 5) - (x^2 - 1)(1)
= 2x^2 + 10x - x^2 + 1
= x^2 + 10x + 1

2. Differentiate the denominator (x+5) as 1.

The derivative of y = (x^2-1)/(x+5) is (x^2 + 10x + 1)/(x+5).

To find the slope of the normal, we can take the negative reciprocal of the derivative at the point (1,0).

1. Substitute x = 1 into the derivative:

m = (1^2 + 10(1) + 1)/(1+5)
= (12)/(6)
= 2

2. Take the negative reciprocal:

Slope of the normal = -1/2

Therefore, the correct answer for the slope of the normal is b) -1/2.

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To find the y-intercept of the line that is tangent to the function f(x) = 4/(3x+1) at the point (-1, -2), we need to find the equation of the tangent line and then determine its y-intercept.

1. Differentiate the function f(x):

f'(x) = d/dx [4/(3x+1)]
= (-4)(3)/(3x+1)^2
= -12/(3x+1)^2

2. Find the slope of the tangent line at x = -1:

m = -12/(3(-1)+1)^2
= -12/(3(-1) + 1)
= -12/(3-1)
= -12/2
= -6

3. Plug the slope (-6) and the point (-1, -2) into the point-slope equation of a line: y - y1 = m(x - x1).

y - (-2) = -6(x - (-1))
y + 2 = -6(x + 1)
y + 2 = -6x - 6
y = -6x - 8

4. Identify the y-intercept of the line (where x = 0):

y = -6(0) - 8
y = -8

Therefore, the correct answer for the y-intercept of the line is d) -2.