Question

the point p(16, 9) lies on the curve ( y = sqrt{x} + 5 ). if q is the point ( (x, sqrt{x} + 5) ), find the slope of the secant line pq for the following values of x. enter your answer as a decimal, be sure to round correctly. if ( x = 16.1 ), the slope of pq is: (\boxed{}) and if ( x = 16.01 ), the slope of pq is: (\boxed{}) and if ( x = 15.9 ), the slope of pq is: (\boxed{}) and if ( x = 15.99 ), the slope of pq is: (\boxed{}) based on the above results, guess the slope of the tangent line to the curve at p(16, 9). (\boxed{})

Explanation:

The slope of the secant line between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by the formula \(m=\frac{y_2 - y_1}{x_2 - x_1}\). Here, \(P(16,9)\) and \(Q(x,\sqrt{x}+5)\), so we will use this formula to calculate the slope for different values of \(x\).

Step 1: Slope formula for secant line \(PQ\)

The slope \(m\) of the line passing through \(P(16,9)\) and \(Q(x,\sqrt{x}+5)\) is given by:
\[m=\frac{(\sqrt{x}+5)-9}{x - 16}=\frac{\sqrt{x}-4}{x - 16}\]
We can simplify the numerator and denominator. Recall that \(a^2-b^2=(a - b)(a + b)\), so \(x-16=(\sqrt{x})^2-4^2=(\sqrt{x}-4)(\sqrt{x}+4)\) (for \(x
eq16\)). Then the slope formula becomes:
\[m=\frac{\sqrt{x}-4}{(\sqrt{x}-4)(\sqrt{x}+4)}=\frac{1}{\sqrt{x}+4}\] (for \(x
eq16\))

Step 2: Calculate slope when \(x = 16.1\)

Substitute \(x = 16.1\) into the simplified slope formula \(m=\frac{1}{\sqrt{x}+4}\)
\(\sqrt{16.1}\approx4.0125\)
\[m=\frac{1}{4.0125 + 4}=\frac{1}{8.0125}\approx0.1248\]

Step 3: Calculate slope when \(x = 16.01\)

Substitute \(x = 16.01\) into the formula \(m=\frac{1}{\sqrt{x}+4}\)
\(\sqrt{16.01}\approx4.00125\)
\[m=\frac{1}{4.00125+4}=\frac{1}{8.00125}\approx0.12498\]

Step 4: Calculate slope when \(x = 15.9\)

Substitute \(x = 15.9\) into the formula \(m=\frac{1}{\sqrt{x}+4}\)
\(\sqrt{15.9}\approx3.9875\)
\[m=\frac{1}{3.9875 + 4}=\frac{1}{7.9875}\approx0.1252\]

Step 5: Calculate slope when \(x = 15.99\)

Substitute \(x = 15.99\) into the formula \(m=\frac{1}{\sqrt{x}+4}\)
\(\sqrt{15.99}\approx3.99875\)
\[m=\frac{1}{3.99875+4}=\frac{1}{7.99875}\approx0.12502\]

Step 6: Guess the slope of the tangent line

As \(x\) approaches \(16\) (from both sides, \(x = 16.1,16.01\) (right - hand side) and \(x = 15.9,15.99\) (left - hand side)), the slope of the secant line approaches \(\frac{1}{\sqrt{16}+4}=\frac{1}{4 + 4}=\frac{1}{8}=0.125\)

Answer:

s:

• When \(x = 16.1\), the slope of \(PQ\) is approximately \(\boldsymbol{0.125}\) (or more precisely \(0.1248\))
• When \(x = 16.01\), the slope of \(PQ\) is approximately \(\boldsymbol{0.125}\) (or more precisely \(0.12498\))
• When \(x = 15.9\), the slope of \(PQ\) is approximately \(\boldsymbol{0.125}\) (or more precisely \(0.1252\))
• When \(x = 15.99\), the slope of \(PQ\) is approximately \(\boldsymbol{0.125}\) (or more precisely \(0.12502\))
• The slope of the tangent line at \(P(16,9)\) is \(\boldsymbol{0.125}\) (or \(\frac{1}{8}\))