Slope of a secant

Question

the point p(16,7) lies on the curve y=sqrt(x) + 3. If q is the point (x, sqrt(x) +3), find the slope of the secant line for pq for the following values. 

x = 16.1 slope of the line is?

x= 16.01 slope of the line is?

 

I have the correct answer but the video its explaining the formula wrong, or im not understanding it. 

formula = f(x₂)-f(x₁) / x₂ - x₁ 

so if i have the points p (16,7) and (16.01 , f(16.01)) shouldnt my equation look something like this..

(fx₂) = sqrt(16.1)+3 

(fx₁) = sqrt(7)+3 // turns out this part is incorrect, i have to use the value 16, i believe 16 should be represented by f₁ shouldnt it be

16 = f₁  and 7 = f(x₁)

16.01 - 16 

 

 

Answer 1

f(x) = √x + 3   f(16) = √16 + 3 = 7

 

slope of pq = [f(x) - f(16)]/(x - 16)

 

                = [(√x+3) - 7]/(x-16) = (√x - 4)/(x - 16)

 

When x = 16.1, we get [f(16.1) - f(16)]/(16.1-16)

 

                                 = (√16.1 - 4)/0.1 = 0.1248

 

When x = 16.01, we have [f(16.01) - f(16)]/(16.01-16)

 

                                    = (√16.01 - 4)/0.01 = 0.12498