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 = - 1, x = 1 a. the equation of the secant line is y = b. the equation of the tangent line is y =

Explanation:

Step1: Find function values at given x - values

For $y = f(x)=x^{2}+x$, when $x=-1$, $y_1=(-1)^{2}+(-1)=1 - 1=0$. When $x = 1$, $y_2=1^{2}+1=2$.

Step2: Calculate slope of secant line

The slope $m_s$ of the secant line passing through $(x_1,y_1)=(-1,0)$ and $(x_2,y_2)=(1,2)$ is $m_s=\frac{y_2 - y_1}{x_2 - x_1}=\frac{2-0}{1-(-1)}=\frac{2}{2}=1$. Using the point - slope form $y - y_1=m(x - x_1)$ with $(x_1,y_1)=(-1,0)$ and $m = 1$, we get $y-0=1\times(x + 1)$, so the equation of the secant line is $y=x + 1$.

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: Calculate slope of tangent line at $x=-1$

When $x=-1$, the slope $m_t$ of the tangent line is $m_t=f^\prime(-1)=2\times(-1)+1=-2 + 1=-1$. Using the point - slope form $y - y_1=m(x - x_1)$ with $(x_1,y_1)=(-1,0)$ and $m=-1$, we get $y-0=-1\times(x + 1)$, so the equation of the tangent line is $y=-x - 1$.

Answer:

a. $y=x + 1$
b. $y=-x - 1$