Question

for the given function, find (a) the equation of the secant line through the points where x has the given values and (b) the equation of the tangent line when x has the first value. y = f(x)=x² + x; x = 3, x = 5 a. the equation of the secant line is y = 9x - 15 b. the equation of the tangent line is y = 7x - 9

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 \(y = 7x-9\).