Y=x^2 -9x at x = 4 Find an equation of the tangent an an equation of the normal

The tangent line to y = f(x) at a point (a, f(a) ) has slope dy/dx at x = a. So first compute the derivative:

y = x² - 9x   →   dy/dx = 2x - 9

When x = 4, the function takes on a value of

y = 4² - 9•4 = -20

and the derivative is

dy/dx (4) = 2•4 - 9 = -1

Then use the point-slope formula to get the equation of the tangent line:

y - (-20) = -1 (x - 4)

y + 20 = -x + 4

y = -x - 24

The normal line is perpendicular to the tangent, so its slope is -1/(-1) = 1. It passes through the same point, so its equation is

y - (-20) = 1 (x - 4)

y + 20 = x - 4

y = x - 24

5/12, or 5: 12

step-by-step explanation: