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:

$m=\frac{y2-y1}{x2-x1}$

When x = 0.3, we have:

$Q(x,\frac{3}{x})=(0.3,\frac{3}{0.3})=(0.3,10)$

Hence, we have:

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

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

Plug in values into the slope formula:

$\begin{gathered} m=\frac{10-15}{0.3-0.2} \\ \\ m=\frac{-5}{0.1} \\ \\ m=-50 \end{gathered}$

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

• If x = 0.21

We have:

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

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

• If x = 0.1:

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

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

• If x = 0.19

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

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