an equation of the line normal to graph of y = x^3 + 3x^2 + 7x - 1at the point x= -1 is?
but y(-1) = -6, so the line is
y + 6 = -1/4 (x + 1)
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)
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 determine the slope of the tangent line. Then, we can find the slope of the normal line by taking the negative reciprocal of the tangent line slope.
Step 1: Find the derivative of the given equation.
The derivative of y = x^3 + 3x^2 + 7x - 1 is:
dy/dx = 3x^2 + 6x + 7
Step 2: Determine the slope of the tangent line.
Substitute x = -1 into the derivative to find the slope at x = -1:
dy/dx = 3(-1)^2 + 6(-1) + 7 = 3 - 6 + 7 = 4
Therefore, the slope of the tangent line at x = -1 is 4.
Step 3: Find the slope of the normal line.
Take the negative reciprocal of the tangent line slope:
Slope of the normal line = -1/4
Step 4: Use the point-slope form to write the equation.
The point-slope form of a line is given by:
y - y1 = m(x - x1)
We are given the point x = -1 and asked to find the equation. Thus, we have:
y - y1 = m(x - x1)
y - (x^3 + 3x^2 + 7x - 1) = -1/4(x - (-1))
y - x^3 - 3x^2 - 7x + 1 = -1/4(x + 1)
This is the equation of the line normal to the graph of y = x^3 + 3x^2 + 7x - 1 at the point x = -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.