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)=x^2+x; x=-1, x=2 (a) The equation of the secant line is y=______ (b) The equation of the tangent line is y=_____ type your answer in slope intercept form

Answer 1

Answer:

Step-by-step explanation:

Given  is a function as

$y=f(x)=x^2+x$

two values of x are x=-1 and x=2

Corresponding y values are

y(-1) = 1-1 =0 and

y(2) = 4+2 =6

Hence the two points are (-1,0) and (2,6)

Secant line equation can be obtained using two point formula for straight lines

a) Secant line joining (-1,0) and (2,6) is

$\frac{y-0}{6-0} =\frac{x+1}{2+1} \\y = 2x+2$\

is the equation of the secant line

b) to find equation of tangent line

Slope of tangent line at (-1,0) = value of derivative at that point

$y' = 2x+1\\y'(-1) = -2+1=-1$

So m =-1 and one point on the line is point of contact (-1,0)

So using point slope formula

$y-0=-1(x+1)\\y = -x-1$