Question

find an equation of the normal line to the parabola y=x^2-5x+4 that is parallel to the line x-3y=5

Answer 1

Slope of the line is 1/3, so we need the point on the parabola where the slope is -3. Take the derivative, set it to -3, solve for x. Find the position of the point on the parabola with that x value. Find the equation of a line with slope 1/3 through that point.

Answer 2

normal line and tangent lines are perpendicular to each other
so the tangent line is perpendicular to x-3y = 5

x-3y = 5 --> y = 1/3x - 5/3
slope = 1/3
perpendicular slope = -3

derivative of f(x) gives the slope at any point
f'(x) = 2x-5
Find x where f'(x) = -3

2x-5 = -3
2x = 2
x=1

f(1) = 0
normal line has slope of 1/3

y = 1/3(x-1)

Answer 3

the last part is where i get confused, the answer is y= (1/3x)-(1/3)

Answer 4

Distribute the 1/3 factor to the (x-1) and you will get that answer.

Answer 5

oh duh! haha sry. but the f(0)=1?

Answer 6

i mean f(1)=0

Answer 7

f(1)=0 means plug in x=1 into the equation for the parabola, and the y value (the value of the function at x=1) is zero.

Answer 8

ohh ok i get it now. thanks