a. use the definition $m_{tan}=lim_{h \to 0}\frac{f(a + h)-f(a)}{h}$ to find the slope of the line tangent…

a. use the definition $m_{tan}=lim_{h \to 0}\frac{f(a + h)-f(a)}{h}$ to find the slope of the line tangent to the graph of $f$ at $p$.\n\nb. determine an equation of the tangent line at $p$.\n\n$f(x)=sqrt{5x + 39}$, $p(5,8)$\n\na. $m_{tan}=square$ (simplify your answer. type an exact answer, using radicals as needed.)

a. use the definition $m_{tan}=lim_{h \to 0}\frac{f(a + h)-f(a)}{h}$ to find the slope of the line tangent to the graph of $f$ at $p$.\n\nb. determine an equation of the tangent line at $p$.\n\n$f(x)=sqrt{5x + 39}$, $p(5,8)$\n\na. $m_{tan}=square$ (simplify your answer. type an exact answer, using radicals as needed.)

Answer

Explanation:

Step1: Identify (a) and (f(x))

Given (f(x)=\sqrt{5x + 39}) and (P(5,8)), so (a = 5).

Step2: Calculate (f(a + h)) and (f(a))

[ \begin{align*} f(a+h)&=\sqrt{5(5 + h)+39}=\sqrt{25+5h + 39}=\sqrt{5h + 64}\ f(a)&=f(5)=\sqrt{5\times5+39}=\sqrt{25 + 39}=8 \end{align*} ]

Step3: Find the limit for the slope (m_{tan})

[ \begin{align*} m_{tan}&=\lim_{h\rightarrow0}\frac{f(a + h)-f(a)}{h}\ &=\lim_{h\rightarrow0}\frac{\sqrt{5h + 64}-8}{h}\ &=\lim_{h\rightarrow0}\frac{(\sqrt{5h + 64}-8)(\sqrt{5h + 64}+8)}{h(\sqrt{5h + 64}+8)}\ &=\lim_{h\rightarrow0}\frac{(5h + 64)-64}{h(\sqrt{5h + 64}+8)}\ &=\lim_{h\rightarrow0}\frac{5h}{h(\sqrt{5h + 64}+8)}\ &=\lim_{h\rightarrow0}\frac{5}{\sqrt{5h + 64}+8}\ &=\frac{5}{\sqrt{64}+8}=\frac{5}{8 + 8}=\frac{5}{16} \end{align*} ]

Step4: Find the equation of the tangent - line

The point - slope form of a line is (y - y_1=m(x - x_1)), where ((x_1,y_1)=(5,8)) and (m=\frac{5}{16}). [ \begin{align*} y-8&=\frac{5}{16}(x - 5)\ y-8&=\frac{5}{16}x-\frac{25}{16}\ y&=\frac{5}{16}x-\frac{25}{16}+8\ y&=\frac{5}{16}x-\frac{25}{16}+\frac{128}{16}\ y&=\frac{5}{16}x+\frac{103}{16} \end{align*} ]

Answer:

a. (m_{tan}=\frac{5}{16}) b. (y=\frac{5}{16}x+\frac{103}{16})