How do you find the equation of the tangent to the curve #y = (4x + 8)/(x + 1)# which is parallel to #y = -4x + 7#?

Answer 1

#y=-4x+8#

If the equation of tangent is parallel to #y=-4x+7#, it has form of #y=-4x+k#
Also, tangent of it must be equal to derivative of #y=(4x+8)/(x+1)# curve
#(4*(x+1)-1*(4x+8))/(x+1)^2=-4#
#-4/(x+1)^2=-4#
#(x+1)^2=1#
#x^2+2x=0#
#x*(x+2)=0#
So, #x_1=-2# and #x_2=0#
For #x=-2#, #y=(4*(-2)+8)/(-2+1)=0/(-1)=0#
For #x=0#, #y=(4*0+8)/(0+1)=8/1=8#

According to equation of tangent formula,

#a)# #y-0=-4*(x-(-2))# or #y=-4x+8# for #(-2, 0)# point.
#b)# #y-8=-4*(x-0)# or #y=-4x+8# for #(0, -2)# point.
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Answer 2

To find the equation of the tangent to the curve y = (4x + 8)/(x + 1) that is parallel to y = -4x + 7, we need to find the derivative of the curve and then use the point-slope form of a line.

First, find the derivative of y = (4x + 8)/(x + 1) using the quotient rule:

dy/dx = [(4)(x + 1) - (4x + 8)(1)] / (x + 1)^2

Simplifying this expression gives:

dy/dx = 4 / (x + 1)^2

Since the tangent line is parallel to y = -4x + 7, its slope will be -4.

Now, set the derivative equal to -4 and solve for x:

4 / (x + 1)^2 = -4

Multiply both sides by (x + 1)^2:

4 = -4(x + 1)^2

Expand and simplify:

4 = -4(x^2 + 2x + 1)

4 = -4x^2 - 8x - 4

Rearrange the equation to standard form:

4x^2 + 8x + 8 = 0

Now, solve this quadratic equation using factoring, completing the square, or the quadratic formula. The solutions will give the x-coordinates of the points where the tangent line intersects the curve.

Once you have the x-coordinate(s), substitute it back into the original equation y = (4x + 8)/(x + 1) to find the corresponding y-coordinate(s).

Finally, use the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is a point on the curve, and m is the slope (-4), to find the equation of the tangent line.

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Answer 3

To find the equation of the tangent to the curve (y = \frac{4x + 8}{x + 1}) that is parallel to the line (y = -4x + 7), we first need to find the derivative of the curve (y = \frac{4x + 8}{x + 1}). Then, we equate this derivative to the slope of the given line, since parallel lines have equal slopes. After finding the value of (x), we substitute it back into the original equation to find the corresponding (y) value. This will give us the point of tangency. Finally, we use the point-slope form of a line to find the equation of the tangent line.

  1. Find the Derivative of the Curve: [y = \frac{4x + 8}{x + 1}] To find the derivative, we can use the quotient rule: [y' = \frac{(4)(x + 1) - (4x + 8)(1)}{(x + 1)^2}] Simplify this expression.

  2. Find the Slope of the Tangent Line: Equate the derivative to the slope of the given line, (y = -4x + 7), since they are parallel. Solve for (x).

  3. Find the Point of Tangency: Substitute the value of (x) found in step 2 back into the original equation to find the corresponding (y) value.

  4. Write the Equation of the Tangent Line: Use the point-slope form of a line: (y - y_1 = m(x - x_1)), where (m) is the slope and ((x_1, y_1)) is a point on the line. Plug in the values of (m), (x_1), and (y_1) found in steps 2 and 3 to get the equation of the tangent line.

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Answer from HIX Tutor

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.

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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