A line perpendicular to another line or to a tangent line is called a normal line. Find an equation of the line perpendicular to the line that is tangent to the following curve at the given point P.
Y=x^2-3x

P (7, 28)

The equation of the normal line at P(7, 28) is:

The equation of the normal line to the curve y = x² - 3x at the point (7, 28) is 11y + x - 315 = 0

Step-by-step explanation:

First, we need to find the slope of the line tangent to the curve at the point (7, 28).

To do this, we need to differentiate y with respect to x and evaluate at the point x0 = 7.

Given y = x² - 3x

dy/dx = 2x - 3

So, the slope, m of the tangent line is dy/dx at x = 7:

m = 2(7) - 3 = 11

Slope, m = 11

The slope, m2 of the line perpendicular to the tangent is given as m2 = -1/m

m2 = −1/11

Finally, given the slope m of a line, and a point (x0, y0) on the line, we can use the point-slope form of the equation of a line:

y - y0=m(x - x0)

The perpendicular line has slope m2 = -1/11, and (7, 28) is a point on that line, the desired equation is:

y - 28 = (-1/11)(x - 7)

Or multiplying by 11, we have

11y - 308 = -x + 7

11y + x - 315 = 0

1/4 because there is 4 grades.