1. The curve y=10/2x+1 -2 intersects the x-axis at A. The tangent to the curve at A intersects the y-axis at C.

(i) Show that the equation of AC is 5y+4x=8
(ii) Find the distance of AC

2. The equation of a curve is y=1/6(2x-3)^3-4x
(i) Find dy/dx
(ii) Find the equation of the tangent to the curve at the point where the curve intersects the y-axis.

the curve intersects the x-axis at (2,0)

y'(x) = -20/(2x+1)^2
y'(2) = -4/5

tangent at (2,0) is
y = -4/5 (x-2)
intersects the y-axis at (0,8/5)

AC is thus
y = (x-2)(-4/5)
5y = 8-4x
5y+4x=8

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y = 1/6 (2x-3)^3 - 4x
y' = (2x-3)^2 - 4

the curve has y-intercept at (0,-9/2)

y'(0) = 5
so, the line is

y+9/2 = 5x

1. (i) To find the x-coordinate where the curve intersects the x-axis, we can set y equal to zero and solve for x:

0 = (10/2x + 1) - 2
2 = 10/2x + 1
1 = 10/2x
1 = 5/x
x = 5

So the curve intersects the x-axis at the point A(5, 0).

Now, let's find the slope of the tangent line at point A. We can find this by taking the derivative of the curve:

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

The slope of the tangent line at point A is -10/5^2 = -2/5.

We can also find the y-intercept of the tangent line. Since the tangent line passes through point A(5, 0), we can substitute these values into the equation of a line:

y = mx + c
0 = (-2/5)(5) + c
0 = -2 + c
c = 2

So the equation of the tangent line AC is y = (-2/5)x + 2.

To check if it's equivalent to 5y + 4x = 8:

Multiply both sides of the equation by 5:
5y + 4x = 2y - 2x + 10

Rearranging:
2y + 2x = 10 (multiply both sides by -1)
-2y - 2x = -10

Adding both equations:
5y + 4x -2y - 2x = 8 -10
3y + 2x = -2

So, the equation of AC is indeed 5y + 4x = 8.

(ii) To find the distance of AC, we can rearrange the equation 5y + 4x = 8 to the standard form, which is Ax + By + C = 0:

5y + 4x = 8
4x + 5y - 8 = 0

Now, we can use the formula for the distance between a point (x₁, y₁) and a line Ax + By + C = 0, which is:

d = |Ax₁ + By₁ + C| / sqrt(A² + B²)

In this case, x₁ = 0 and y₁ = 0 (as point C is the y-intercept of the line AC):

d = |4(0) + 5(0) - 8| / sqrt(4² + 5²)
= |0 - 8| / sqrt(16 + 25)
= |-8| / sqrt(41)
= 8 / sqrt(41)

So, the distance of AC is 8 / sqrt(41).

2. (i) To find dy/dx, we need to use the power rule for differentiation. The power rule states that if y = axⁿ, then dy/dx = anxⁿ⁻¹.

In this case, y = (1/6)(2x - 3)³ - 4x. Let's differentiate:

dy/dx = (1/6)(3)(2x - 3)²(2) - 4
= (1/3)(2x - 3)² - 4

So, dy/dx = (1/3)(2x - 3)² - 4.

(ii) To find the point of intersection with the y-axis, we can set x equal to zero and solve for y:

y = (1/6)(2(0) - 3)³ - 4(0)
= (1/6)(-3)³
= (1/6)(-27)
= -27/6
= -9/2

So, the curve intersects the y-axis at the point (0, -9/2).

Now, we can find the slope of the tangent line at this point by substituting x = 0 into the derivative we found in part (i):

dy/dx = (1/3)(2(0) - 3)² - 4
= (1/3)(-3)² - 4
= (1/3)(9) - 4
= 3 - 4
= -1

The slope of the tangent line at the point of intersection with the y-axis is -1.

Using the point-slope form of a line, we can write the equation of the tangent line:

y - (-9/2) = -1(x - 0)
y + 9/2 = -x

Multiply both sides by 2:
2y + 9 = -2x

Rearranging the terms:
2y + 2x = -9

So, the equation of the tangent line to the curve at the point of intersection with the y-axis is 2y + 2x = -9.

Sure! Let's solve the given questions step by step:

1. (i) To show that the equation of AC is 5y + 4x = 8, we need to find the coordinates of the points A and C.

The given equation of the curve is y = (10/2)x + 1 - 2. We can simplify it as follows:

y = 5x + 1 - 2
y = 5x - 1

Now, to find point A, we set y = 0 (since A lies on the x-axis):

0 = 5x - 1
5x = 1
x = 1/5

So, point A is (1/5, 0).

Next, we need to find the equation of the tangent to the curve at point A. To find the slope of the tangent at A, we need to find the derivative of the curve equation with respect to x:

dy/dx = d/dx(5x - 1)
= 5

The slope of the tangent at A is 5. Since the tangent passes through point A (1/5, 0), we can use the point-slope form of a line to find the equation of the tangent:

y - 0 = 5(x - 1/5)
5y = 5x - 1
5y - 5x = -1
5y + 4x = 8

Therefore, the equation of AC is 5y + 4x = 8.

(ii) To find the distance of AC, we need to find the coordinates of point C.

To find point C, we set x = 0 (since C lies on the y-axis):

5y + 4(0) = 8
5y = 8
y = 8/5

So, point C is (0, 8/5).

Now, we can use the distance formula to find the distance between points A and C:

Distance of AC = √[(x₂ - x₁)² + (y₂ - y₁)²]
= √[(0 - 1/5)² + (8/5 - 0)²]
= √[(1/5)² + (8/5)²]
= √[(1/25) + (64/25)]
= √(65/25)
= √(13/5)
= √13/√5
= (√13/√5) * (√5/√5)
= √65/5

Therefore, the distance of AC is (√65)/5.

2. (i) To find dy/dx for the curve y = (1/6)(2x - 3)³ - 4x, we need to find the derivative of the curve equation with respect to x:

dy/dx = d/dx[(1/6)(2x - 3)³ - 4x]
= (1/6) * d/dx(2x - 3)³ - d/dx(4x)
= (1/6) * 3(2x - 3)² * 2 - 4
= (1/6) * 6(2x - 3)² - 4
= (2x - 3)² - 4

Therefore, dy/dx = (2x - 3)² - 4.

(ii) To find the equation of the tangent to the curve at the point where it intersects the y-axis, we need to find the y-intercept of the tangent.

To find the y-intercept, we substitute x = 0 into the curve equation:

y = (1/6)(2(0) - 3)³ - 4(0)
= (1/6)(-3)³
= (1/6)(-27)
= -27/6
= -9/2

So, the point of intersection with the y-axis is (0, -9/2).

Since the point (0, -9/2) lies on the tangent line, we can use the point-slope form of a line to find the equation of the tangent:

y - (-9/2) = (2x - 3)² - 4(0)
y + (9/2) = (2x - 3)²
y = (2x - 3)² - 9/2

Therefore, the equation of the tangent to the curve at the point where it intersects the y-axis is y = (2x - 3)² - 9/2.