Question

The point P(16, 6) lies on the curve y = sqrt(x) + 2 If Q is the point (x, sqrt(x) + 2) , find the slope of the secant line PQ for the following values of

Answer 1

Given:

`P(16,6)\text{ and Q \lparen x,}√(x)+2)`

Required:

`We\text{ need to find the slope of the secant line PQ.}`

Step-by-step explanation:

Consider the slope of the secant formula.

`Slopw=(y_2-y_1)/(x_2-x_1)`

1)

Substitute x =16.1 in the point Q.

`\text{ Q \lparen x,}√(x)+2)=Q(16.1,√(16.1)+2)`
`\text{ Q \lparen x,}√(x)+2)=Q(16.1,6.0125)`
`Substitute\text{ }x_1=16,x_2=16.1,y_1=6,\text{ and }y_2=6.0125\text{ in the slope formula.}`
`Slope=(6.0125-6)/(16.1-16)`
`Slope=(0.0125)/(0.1)`

`Slope=0.125`

2)

Substitute x =16.01 in the point Q.

`\text{ Q \lparen x,}√(x)+2)=Q(16.01,√(16.01)+2)`
`\text{ Q \lparen x,}√(x)+2)=Q(16.01,6.00125)`
`Substitute\text{ }x_1=16,x_2=16.01,y_1=6,\text{ and }y_2=6.00125\text{ in the slope formula.}`
`Slope=(6.00125-6)/(16.01-16)`
`Slope=(0.00125)/(0.01)`

`Slope=0.125`

3)

Substitute x =15.9 in the point Q.

`\text{ Q \lparen x,}√(x)+2)=Q(15.9,√(15.9)+2)`
`\text{ Q \lparen x,}√(x)+2)=Q(15.9,5.9875)`
`Substitute\text{ }x_1=16,x_2=15.9y_1=6,\text{ and }y_2=5.9875\text{ in the slope formula.}`
`Slope=(5.9875-6)/(15.9-16)`
`Slope=(-0.0125)/(-0.1)`
`Slope=0.125`

4)

Substitute x =15.99 in the point Q.

`\text{ Q \lparen x,}√(x)+2)=Q(15.99,√(15.99)+2)`
`\text{ Q \lparen x,}√(x)+2)=Q(15.99,5.9987)`
`Substitute\text{ }x_1=16,x_2=15.99y_1=6,\text{ and }y_2=5.9987\text{ in the slope formula.}`
`Slope=(5.9987-6)/(15.99-16)`
`Slope=(-0.00125)/(-0.01)`
`Slope=(-0.00125)/(-0.01)`
`Slope=0.125`

Final answer:

The slope of the tangent line PQ is

`Slope=0.12\frac{}{}`