Find The X Co-ordinate.

Question

Find the x co-ordinate of the point where the normal to f(x) = x² − 3x + 1 at x = −1 intersects the curve again.

Answer 1

y = x²-3x+1 (1)

at x=-1, y=(-1)²-3(-1)+1 = 1+3+1 = 5 --> the curve passes through (-1,5). Will use this info shortly.

To find the equation of the normal line, we first need the slope of the tangent line at x=-1:

dy/dx = 2x-3 --> at x=-1 y = 2(-1)-3 = -5

The slope of the normal line (lines that are perpendicular have slopes that are negative reciprocals of each other) is therefore -1/(-5) = 1/5

We know the normal line meets the curve at (-1,5) so the equation of the normal line can be written using point-slope form of a line:

y_n-5 = (1/5)(x_n + 1) where I am using subscript _n to differentiatate from the original curve.

Re-writing the normal line in slope-intercept form (basically, solve for y_n):

y_n = (1/5)x_n + (26/5) (2)

At the other point of intersection with (1) y_n will equal y. We set the RHS of (1) equal to the RHS of (2):

(note how I can use x or x_n at this point because they are equal)

(1/5)x + (26/5) = x²-3x+1

Multiply both sides by 5:

x + 26 = 5x²-15x+5

Re-arrange:

5x²-16x-21 = 0

Factor:

(x+1)(5x-21) = 0

Solutions are x=-1 (we already have that one), and x = 21/5

If you wanted to also know the y value you can plug in x=21/5 into (1) (I get 6.04)