Question

Question attached and investigates relationship between zeroes and tangents of cubic function

Answer 1

The tangent lines of the cubic function f(x) = x(x - 2)(x - 6) at the zeros 0, 2 and 6 are y = 12x, y = -8x + 16 and y = 24x - 144

Part A: The tangent lines of f(x) = x(x - 2)(x - 6)

The cubic function is given as:

f(x) = x(x - 2)(x - 6)

Expand

f(x) = x³ - 8x² + 12x

Differentiate the function

m = 3x² - 16x + 12

At the point x = 0, we have:

m = 3(0)² - 16(0) + 12

m = 12

Also, we have:

f(0) = 0

A linear equation is represented as:

y = m(x - x₁) + y₁

So, we have:

y = 12(x - 0) + 0

y = 12x

At the point x = 2, we have:

m = 3(2)² - 16(2) + 12

m = -8

Also, we have:

f(2) = 0

A linear equation is represented as:

y = m(x - x₁) + y₁

So, we have:

y = -8(x - 2) + 0

y = -8x + 16

At the point x = 6, we have:

m = 3(6)² - 16(6) + 12

m = 24

Also, we have:

f(6) = 0

A linear equation is represented as:

y = m(x - x₁) + y₁

So, we have:

y = 24(x - 6) + 0

y = 24x - 144

The above means that the tangent lines are:

y = 12x, y = -8x + 16 and y = 24x - 144

From the attached graph, we can see that the tangent lines at the zero values x = 0 and x = 2 intersect the graph of f(x) at the zero value x = 6

Part B: The tangent lines of f(x) = (x - a)(x - b)(x - c)

The equation is given as:

f(x) = (x - a)(x - b)(x - c)

To prove the statement using a computer algebra system, simply follow the next steps:

• Set values for a, b and c
• Include the following vertices in the plane: (a,f(a)), (b,f(b)) and (c,f(c))
• Differentiate f(x) i.e g(x) = f'(x)
• Include the following vertices in the plane: (a,g(a)), (b,g(b)) and (c,g(c))
• Lastly, plot the following equations: y = g(a)(x - a) + f(a), y = g(b)(x - b) + f(b) and y = g(c)(x - c) + f(c)

See attachment (2) for proof

Read more about tangent lines at:

brainly.com/question/6617153

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