Given:
[tex]P(16,6)\text{ and Q \lparen x,}\sqrt{x}+2)[/tex]Required:
[tex]We\text{ need to find the slope of the secant line PQ.}[/tex]Explanation:
Consider the slope of the secant formula.
[tex]Slopw=\frac{y_2-y_1}{x_2-x_1}[/tex]1)
Substitute x =16.1 in the point Q.
[tex]\text{ Q \lparen x,}\sqrt{x}+2)=Q(16.1,\sqrt{16.1}+2)[/tex][tex]\text{ Q \lparen x,}\sqrt{x}+2)=Q(16.1,6.0125)[/tex][tex]Substitute\text{ }x_1=16,x_2=16.1,y_1=6,\text{ and }y_2=6.0125\text{ in the slope formula.}[/tex][tex]Slope=\frac{6.0125-6}{16.1-16}[/tex][tex]Slope=\frac{0.0125}{0.1}[/tex][tex]Slope=0.125[/tex]2)
Substitute x =16.01 in the point Q.
[tex]\text{ Q \lparen x,}\sqrt{x}+2)=Q(16.01,\sqrt{16.01}+2)[/tex][tex]\text{ Q \lparen x,}\sqrt{x}+2)=Q(16.01,6.00125)[/tex][tex]Substitute\text{ }x_1=16,x_2=16.01,y_1=6,\text{ and }y_2=6.00125\text{ in the slope formula.}[/tex][tex]Slope=\frac{6.00125-6}{16.01-16}[/tex][tex]Slope=\frac{0.00125}{0.01}[/tex][tex]Slope=0.125[/tex]3)
Substitute x =15.9 in the point Q.
[tex]\text{ Q \lparen x,}\sqrt{x}+2)=Q(15.9,\sqrt{15.9}+2)[/tex][tex]\text{ Q \lparen x,}\sqrt{x}+2)=Q(15.9,5.9875)[/tex][tex]Substitute\text{ }x_1=16,x_2=15.9y_1=6,\text{ and }y_2=5.9875\text{ in the slope formula.}[/tex][tex]Slope=\frac{5.9875-6}{15.9-16}[/tex][tex]Slope=\frac{-0.0125}{-0.1}[/tex][tex]Slope=0.125[/tex]4)
Substitute x =15.99 in the point Q.
[tex]\text{ Q \lparen x,}\sqrt{x}+2)=Q(15.99,\sqrt{15.99}+2)[/tex][tex]\text{ Q \lparen x,}\sqrt{x}+2)=Q(15.99,5.9987)[/tex][tex]Substitute\text{ }x_1=16,x_2=15.99y_1=6,\text{ and }y_2=5.9987\text{ in the slope formula.}[/tex][tex]Slope=\frac{5.9987-6}{15.99-16}[/tex][tex]Slope=\frac{-0.00125}{-0.01}[/tex][tex]Slope=\frac{-0.00125}{-0.01}[/tex][tex]Slope=0.125[/tex]Final answer:
The slope of the tangent line PQ is
[tex]Slope=0.12\frac{}{}[/tex]