an equation of the line normal to graph of y = x^3 + 3x^2 + 7x - 1at the point x= -1 is?

Answer 1

To find the equation of the line normal to the graph of y = x^3 + 3x^2 + 7x - 1 at the point x = -1, we need to find the slope of the tangent line at x = -1 first.

The slope of the tangent line is equal to the derivative of y with respect to x evaluated at x = -1.

Taking the derivative of y = x^3 + 3x^2 + 7x - 1:
dy/dx = 3x^2 + 6x + 7

Evaluating the derivative at x = -1:
dy/dx = 3(-1)^2 + 6(-1) + 7
dy/dx = 3 - 6 + 7
dy/dx = 4

The slope of the tangent line at x = -1 is 4.

Since the line normal to a curve has a slope that is the negative reciprocal of the slope of the tangent line, the slope of the line normal to the graph at x = -1 is -1/4.

Now we have the slope of the line and a point on the line (x = -1, y = -2), we can use the point-slope form of a line to find the equation:

y - y₁ = m(x - x₁)

Substituting the values:
y - (-2) = -1/4(x - (-1))
y + 2 = -1/4(x + 1)

Simplifying:
y + 2 = -1/4x - 1/4
y = -1/4x - 1/4 - 2
y = -1/4x - 1/4 - 8/4
y = -1/4x - 9/4

Therefore, the equation of the line normal to the graph of y = x^3 + 3x^2 + 7x - 1 at the point x = -1 is y = -1/4x - 9/4.

Answer 2

but y(-1) = -6, so the line is
y + 6 = -1/4 (x + 1)

Answer 3

Apologies for the mistake. You are correct.

Given that y(-1) = -6, the equation of the line normal to the graph of y = x^3 + 3x^2 + 7x - 1 at the point x = -1 is:

y + 6 = -1/4 (x + 1)