for the given function, find (a) the equation of the secant line through the points where x has the given…

Explanation:

Step1: Find function values at given x - values

For (y = f(x)=x^{2}+x), when (x = 3), (y_1=f(3)=3^{2}+3=9 + 3=12); when (x = 5), (y_2=f(5)=5^{2}+5=25 + 5=30).

Step2: Calculate slope of secant line

The slope (m) of the secant line passing through ((x_1,y_1)=(3,12)) and ((x_2,y_2)=(5,30)) is (m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{30 - 12}{5 - 3}=\frac{18}{2}=9). Using the point - slope form (y - y_1=m(x - x_1)) with ((x_1,y_1)=(3,12)) and (m = 9), we get (y-12=9(x - 3)), which simplifies to (y=9x-15).

Step3: Find derivative of the function

The derivative of (y = f(x)=x^{2}+x) using the power rule ((x^n)^\prime=nx^{n - 1}) is (y^\prime=f^\prime(x)=2x + 1).

Step4: Find slope of tangent line at (x = 3)

Substitute (x = 3) into the derivative: (m_{tangent}=f^\prime(3)=2\times3+1=7). Using the point - slope form (y - y_1=m(x - x_1)) with ((x_1,y_1)=(3,12)) and (m = 7), we get (y-12=7(x - 3)), which simplifies to (y=7x-9).

Answer:

a. The equation of the secant line is (y = 9x-15). b. The equation of the tangent line is (y = 7x-9).