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

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.\ny = f(x)=x² + x; x = - 3, x = 1

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.\ny = f(x)=x² + x; x = - 3, x = 1

Answer

Explanation:

Step1: Find the y - values of the two points

For (x=-3), (y = f(-3)=(-3)^2+( - 3)=9 - 3=6). For (x = 1), (y=f(1)=1^2 + 1=2). So the two points are ((-3,6)) and ((1,2)).

Step2: Find the slope of the secant line

The slope (m_s) of the secant line passing through ((x_1,y_1)=(-3,6)) and ((x_2,y_2)=(1,2)) is (m_s=\frac{y_2 - y_1}{x_2 - x_1}=\frac{2 - 6}{1-(-3)}=\frac{-4}{4}=-1). Using the point - slope form (y - y_1=m(x - x_1)) with the point ((1,2)), the equation of the secant line is (y - 2=-1(x - 1)), which simplifies to (y=-x + 3).

Step3: Find the derivative of the function

The derivative of (y=f(x)=x^2+x) using the limit definition (\lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}) (as shown in the image (\lim_{h\rightarrow0}\frac{[(x + h)^2+(x + h)]-(x^2+x)}{h}=\lim_{h\rightarrow0}(2x+h + 1)=2x + 1)).

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

Substitute (x=-3) into the derivative. The slope (m_t) of the tangent line at (x=-3) is (m_t=2(-3)+1=-6 + 1=-5).

Step5: Find the equation of the tangent line

Using the point - slope form (y - y_1=m(x - x_1)) with the point ((-3,6)) and (m=-5), we have (y - 6=-5(x+3)). Expanding gives (y-6=-5x-15), and simplifying gives (y=-5x - 9).

Answer:

(a) The equation of the secant line is (y=-x + 3). (b) The equation of the tangent line is (y=-5x - 9).