What is the equation of the normal line to the curve y = x^2 - 3x + 5

Here's the whole problem:

Find a equation of the normal line to the curve y = x^2 - 3x + 5 at the point where the slope of the tangent line and the rate of change of the slope of the tangent line are equal.

Thanks! (:

The slope of the tangent line for all x is the first derivative of y, which may be denoted as y'.

So, y' = 2x - 3.

The rate of change of the slope of the tangent line is the second derivative of y, which may be denoted as y''.

So, y'' = 2.

We equate y' and y'' since they should be equal, as given in the question.

2x - 3 = 2
x = 5/2

Therefore, at x = 5/2, the slope of the tangent line is 2. Since we are looking for the normal line, we need the line perpendicular to the tangent. Hence, the slope of the normal should be the negative reciprocal of the slope of the tangent. So, the negative reciprocal of 2 is -1/2, and that's the slope of the normal line.

We now calculate for y when x = 5/2.

y = (5/2)^2 - 3(5/2) + 5 = 15/4

There we have it. We now have all we need to find the equation of the normal line.

y - 15/4 = -1/2(x - 5/2)

Finally, after simplifying, we get 4y + 2x - 20 = 0. That is the equation of the normal line.

y = x^2 - 3x + 5
y' = 2x - 3
2x - 3 = 2
2x = 5
x = 5/2
y(5/2) = 15/4
y'(5/2) = 2
slope of the normal = -1/2
y - 15/4 = -1/2(x - 5/2)
y = -1/2*x + 5

y=x^2-3x+5 ,dy/dx= 2x-3, d^2y/dx^2=2, 2x-3=2, x=5/2, y=15/4