Derivative problem, f(x)=x^2+x, to find the equation of the secant line.

Question

Find the equation of the secant line through the points where x has the given values and the equation of the tangent line when x has the first value, x=-1 and x=3

Answer 1

Calculating the secant line is quite simple, we first need to find the y-values for the function at x = -1 and x = 3.

f(-1) = (-1)² + (-1) = 1 - 1 = 0

f(3) = 3² + 3 = 9 + 3 = 12

From here, we know that the secant line between any 2 point on a curve is a linear function, so we just need to plug values into the linear equation formula from algebra to get our answer.

y - y₁ = m * (x - x₁)

In this equation, the point (x₁, y₁) can be any point that is on the linear function. For this problem, I will choose the point (-1, f(-1)), but you could choose the point (3, f(3)) to get the same linear function in the end. We will use the latter to check out answer. Also, in this function, m represents the slope of the line, so we will need to use the slope formula from algebra.

m = (y₂ - y₁)/(x₂ - x₁₎

_{Set the points} (x₁, y₁) = (-1, f(-1)) = (-1, 0) and (x₂, y₂) = (3, f(3)) = (3, 12)

m = (12 - 0)/(3 - (-1)) = 12 / 4 = 3

Plugging in to the linear equation formula

y - 0 = 3 (x - (-1))

y = 3x + 3

Checking our equation for both points

0 = 3(-1) + 3

12 = 3(3) + 3

So y = 3x + 3 is the secant line.

For the tangent line on the curve at x = -1, we can apply the same general formula for a linear function here as well, using the point (-1,0). We just need to find the slope of the function at x = -1, which we can find by evaluating the derivative f(x) at x = -1.

f'(x) = 2x + 1

f'(-1) = -2 + 1 = -1 = m

Plugging this into the linear equation formula

y - 0 = -1(x - (-1))

y = -x - 1

Plugging in (-1, 0) to see if the linear function passes through the point

0 = -(-1) - 1 = 1 - 1 = 0

So y = -x - 1 is indeed the tangent line at x = -1