Question

Slope of tangent and secant line?

Answer 1

(a).

The function is f(x) = x² - 8x.

If x = 1 ⇒ f(1) = (1)² - 8(1) = -7.

If x = 9 ⇒ f(9) = (9)² - 8(9) = 9.

Find the slope of secant line:

Substitute (x₁, y₁) = (1, -7) and (x₂, y₂) = (9, 9) in the slope formula: m = (y₂ - y₁)/(x₂ - x₁).

m = (9 + 7)/(9 - 1) = 16/8 = 2.

So the slope of secant line = 2

Answer 2

(b).

The function is f(x) = x² - 8x.

If x = 5 ⇒ f(5) = (5)² - 8(5) = 25 - 40 = -15.

If x = 5+h ⇒ f(5+h) = (5+h)² - 8(5+h) = (5+h)[(5+h) - 8] = (5+h)(h-3) = h²+2h-15.

Find the slope of secant line:

Substitute (x₁, y₁) = (5, -15) and (x₂, y₂) = (5+h, h²+2h-15) in the slope formula: m = (y₂ - y₁)/(x₂ - x₁).

m = (h²+2h-15+15)/(5+h-5) = (h²+2h)/(h) = h+2.

So the slope of secant line = h + 2.

Answer 3

(c).

The function is f(x) = x² - 8x.

If x = 5 ⇒ f(5) = (5)² - 8(5) = 25 - 40 = -15.

Differentiate with respect to x.

f'(x) = 2x - 8

The slope of tangent line at the point (5, -15).

m = f'(5) = 2(5) - 8 = 2.

The slope of tangent line is 2.

Answer 4

(d).

The function is f(x) = x² - 8x.

If x = 5 ⇒ f(5) = (5)² - 8(5) = 25 - 40 = -15.

Differentiate with respect to x.

f'(x) = 2x - 8

The slope of tangent line at the point (5, -15).

m = f'(5) = 2(5) - 8 = 2.

The slope of tangent line is 2.

Substitute (x₁, y₁) = (5, -15) and m = 2 in the point-slope form of line equation: y - y₁ = m(x - x₁).

y + 15 = 2(x - 5)

y + 15 = 2x - 10

y = 2x - 25.

The tangent line equation is y = 2x - 25.