Final answer:
The equation of the line normal to f(x) = 2x² - 3x at the point (1, -1) is found by calculating the derivative to get the slope of the tangent line, finding the negative reciprocal to obtain the normal line slope, and then using the point-slope form to find the equation, which is y = -x.
Step-by-step explanation:
To find the equation of the line normal to the curve of f(x) = 2x² - 3x at the point (1, -1), we must first find the slope of the tangent to the curve at that point. The slope of the tangent is the derivative of f(x) evaluated at the point x=1.
First, we find the derivative:
f'(x) = d/dx (2x² - 3x) = 4x - 3
Then, we evaluate the derivative at x=1:
f'(1) = 4(1) - 3 = 1
The slope of the normal line is the negative reciprocal of the slope of the tangent, hence the slope of the normal is -1 (since the negative reciprocal of 1 is -1).
The equation of a line in point-slope form is:
y - y1 = m(x - x1)
Using the point (1, -1) and the slope -1:
y - (-1) = -1(x - 1)
y + 1 = -x + 1
Subtracting 1 from both sides, we get:
y = -x
The equation of the line normal to f(x) at the point (1, -1) is y = -x.