(1 point)\nthe point $p(16,6)$ lies on the curve $y = \\sqrt{x}+2$. let $q$ be the point $(x,\\sqrt{x}+2)$.\n…

(1 point)\nthe point $p(16,6)$ lies on the curve $y = \\sqrt{x}+2$. let $q$ be the point $(x,\\sqrt{x}+2)$.\na.) find the slope of the secant line $pq$ for the following values of $x$. (answers here should be correct to at least 6 places after the decimal point.)\nif $x = 16.1$, the slope of $pq$ is:\nif $x = 16.01$, the slope of $pq$ is:\nif $x = 15.9$, the slope of $pq$ is:\nif $x = 15.99$, the slope of $pq$ is:\nb.) based on the above results, estimate the slope of the tangent line to the curve at $p(16,6)$.\nanswer:

(1 point)\nthe point $p(16,6)$ lies on the curve $y = \\sqrt{x}+2$. let $q$ be the point $(x,\\sqrt{x}+2)$.\na.) find the slope of the secant line $pq$ for the following values of $x$. (answers here should be correct to at least 6 places after the decimal point.)\nif $x = 16.1$, the slope of $pq$ is:\nif $x = 16.01$, the slope of $pq$ is:\nif $x = 15.9$, the slope of $pq$ is:\nif $x = 15.99$, the slope of $pq$ is:\nb.) based on the above results, estimate the slope of the tangent line to the curve at $p(16,6)$.\nanswer:

Answer

Explanation:

Step1: Recall slope - formula

The slope of the secant line passing through two points $(x_1,y_1)$ and $(x_2,y_2)$ is $m=\frac{y_2 - y_1}{x_2 - x_1}$. Here, $P(16,6)$ so $x_1 = 16,y_1=6$, and $Q(x,\sqrt{x}+2)$ so $x_2=x,y_2=\sqrt{x}+2$. Then the slope of the secant line $PQ$ is $m=\frac{\sqrt{x}+2 - 6}{x - 16}=\frac{\sqrt{x}-4}{x - 16}$.

Step2: When $x = 16.1$

Substitute $x = 16.1$ into the slope formula: [ \begin{align*} m&=\frac{\sqrt{16.1}-4}{16.1 - 16}\ &=\frac{\sqrt{16.1}-4}{0.1}\ &\approx\frac{4.01248 - 4}{0.1}\ &=\frac{0.01248}{0.1}\ &= 0.124800 \end{align*} ]

Step3: When $x = 16.01$

Substitute $x = 16.01$ into the slope formula: [ \begin{align*} m&=\frac{\sqrt{16.01}-4}{16.01 - 16}\ &=\frac{\sqrt{16.01}-4}{0.01}\ &\approx\frac{4.00125 - 4}{0.01}\ &=\frac{0.00125}{0.01}\ &=0.125000 \end{align*} ]

Step4: When $x = 15.9$

Substitute $x = 15.9$ into the slope formula: [ \begin{align*} m&=\frac{\sqrt{15.9}-4}{15.9 - 16}\ &=\frac{\sqrt{15.9}-4}{-0.1}\ &\approx\frac{3.98748 - 4}{-0.1}\ &=\frac{-0.01252}{-0.1}\ &=0.125200 \end{align*} ]

Step5: When $x = 15.99$

Substitute $x = 15.99$ into the slope formula: [ \begin{align*} m&=\frac{\sqrt{15.99}-4}{15.99 - 16}\ &=\frac{\sqrt{15.99}-4}{-0.01}\ &\approx\frac{3.99875 - 4}{-0.01}\ &=\frac{-0.00125}{-0.01}\ &=0.125000 \end{align*} ]

Step6: Estimate the slope of the tangent line

As $x$ gets closer to 16 from both the left - hand side ($x=15.9,15.99$) and the right - hand side ($x = 16.01,16.1$), the slopes of the secant lines approach $0.125000$. So the estimated slope of the tangent line at $P(16,6)$ is $0.125000$.

Answer:

When $x = 16.1$, the slope of $PQ$ is $0.124800$. When $x = 16.01$, the slope of $PQ$ is $0.125000$. When $x = 15.9$, the slope of $PQ$ is $0.125200$. When $x = 15.99$, the slope of $PQ$ is $0.125000$. The estimated slope of the tangent line at $P(16,6)$ is $0.125000$.