Equation of the Secant Line...2

[nycmathguy]

Find the equation of the secant line given that
x = 1 and x = 2.

f(x) = x^2 - 4

Evaluate f(x) at x = 1 and x = 2.

f(1) = (1)^2 - 4

f(1) = 1 - 4

f(1) = -3

One point is (1, -3).

f(2) = (2)^2 - 4

f(2) = 4 - 4

f(2) = 0

Another point is (2, 0).

Use (1, -3) to form a linear equation.

y = mx + b_1

-3 = m(1) + b_1

-3 = m + b_1

-3 - m = b_1

Use (2, 0) to form another linear equation.

y = mx + b_2

0 = m(2) + b_2

0 = 2m + b_2

0 - 2m = b_2

-2m = b_2

Now set b_1 = b_2 to find m.

-m - 3 = -2m

-3 = -2m + m

- 3 = -m

-3/-1 = m

3 = m

I now have y = 3x + b.

Use (1, -3) or (2, 0) to find b.

I will use (2, 0).

y = 3x + b

0 = 3(2) + b

0 = 6 + b

0 - 6 = m

-6 = m. . .the slope of the secant line we seek.

So, the equation of the secant line is y = 3x - 6.

You say?

 

[MathLover1]

correct

 

[nycmathguy]

correct

I like it. I love it. I need more calculus.