The points P({4},{1}) and Q(x,y) are on the graph of the function f(x) = {\sqrt{x}-1}.

Question

The points P(4, 1) and Q(x, y) are on the graph of the function f(x) = √x - 1. Complete the table with the values for the slope of the secant line passing through points P and Q. Do not round until the end of your calculations, then round to 5 decimal places.

Answer 1

f(x) = sqr(x) -1, graphically is the top half of a rightward opening parabola. with vertex (0,-1)

as x approaches 4, the slope of the secant line approaches the slope of the tangent line at (4,1)

which equals the derivative f'(x) = 1/2sqrx

where f'(4) = 1/2sqr4 = 1/2(2) = 1/4 = 0.25

the last line of the table should be very near 0.25

the lines above it getting slightly smaller

a rough sketch of the graph will show the slope getting smaller as x gets larger

the 1st line of the table has slope of the secant line drawn from (4,1) to (x,y) = (4,1) to (4.1, f(4.1))

which

= change in y coordinates divided by change in x coordinates

=(y- 1)/(x-4)

= ((f(4.1) -1)/(4.1-4)

= (sqr4.1-1-1)/0.1

= 10(sqr4.1-2)

=10(2.024846-2)

=10(.024846)

=.24845

2nd and 3rd rows will have values between .24845 and .25000

calculate them the same way

calculate the last row the same way, but expect a value very close to .25

an educated guess, estimate or interpolation might be about .24999,

2nd row about .24900, 3rd row about .24950