Question

I have a calculus question about points on a second line. pic included

Answer 1

Given:

P(¹/₅,15) lies on the curve y = 3/x

Q(x, 3/x)

Let's find the slope for the following values of x:

• If x = 0.3:

Apply the slope formula:

`m=(y2-y1)/(x2-x1)`

When x = 0.3, we have:

`Q(x,(3)/(x))=(0.3,(3)/(0.3))=(0.3,10)`

Hence, we have:

(x1, y1) ==> (1/5, 15) ==> (0.2, 15)

(x2, y2) ==> (0.3, 10)

Plug in values into the slope formula:

`\begin{gathered} m=(10-15)/(0.3-0.2) \\ \\ m=(-5)/(0.1) \\ \\ m=-50 \end{gathered}`

If x = 0.3, the slope of PQ is -50.

• If x = 0.21

We have:

`\begin{gathered} Q(x,(3)/(x))=(0.21,(3)/(0.21))=(0.21,\text{ 14.2857\rparen} \\ \\ \text{ The slope wil be:} \\ m=(14.2857-15)/(0.21-0.2) \\ \\ m=-(0.714)/(0.01)=-71.4 \end{gathered}`

If x = 0.21, the slope of PQ is -71.4

• If x = 0.1:

`\begin{gathered} Q(x,(3)/(x))==>(0.1,(3)/(0.1))==>(0.1,30) \\ \\ m=(30-15)/(0.1-0.2)=(15)/(-0.1)=-150 \end{gathered}`

If x = 0.1, the slope of PQ is -150

• If x = 0.19

`\begin{gathered} Q(x,(3)/(x))==>Q(0.19,(3)/(0.19))==>(0.19,15.789) \\ \\ m=(15.789-15)/(0.19-0.2)=-78.9 \end{gathered}`

If x = 0.19, the slope of PQ = -78.9

• Part B.

Based on the above results, the slope of the tangent line to the curve at P(0.2, 15) will be between: -71.4 to -78.9

Therefore, the slope will be -75

ANSWER:

If x = 0.3, the slope of PQ is -50.

If x = 0.21, the slope of PQ is -71.4

If x = 0.1, the slope of PQ is -150

If x = 0.19, the slope of PQ = -78.9

Part B.

The predicted slope of the tangent line to the curve is -75