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 = 3 a. the equation of the secant line is y = 5x - 3 b. the equation of the tangent line is y = □

Explanation:

Step1: Find the derivative of the function

The function is $y = f(x)=x^{2}+x$. Using the power - rule, if $y = x^{n}$, then $y^\prime=nx^{n - 1}$. So $y^\prime=f^\prime(x)=2x + 1$.

Step2: Evaluate the derivative at $x = 1$

When $x = 1$, $f^\prime(1)=2\times1+1=3$. This is the slope of the tangent line at $x = 1$.

Step3: Find the value of the function at $x = 1$

When $x = 1$, $y=f(1)=1^{2}+1=2$.

Step4: Use the point - slope form of a line

The point - slope form of a line is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})$ is a point on the line and $m$ is the slope. Here, $x_{1}=1$, $y_{1}=2$ and $m = 3$. So $y-2=3(x - 1)$.

Step5: Simplify the equation

Expand the right - hand side: $y-2=3x-3$. Then, $y=3x - 1$.

Answer:

$y = 3x-1$