Respuesta :

apologiabiology
find the line that is normal to the parabola at the given point
remember that normal means perpendicular
perpendicular lines have slopes that multiply to -1
we can use point slope form to write the equation of the line since we are given the point (1,0)

we just need the slope

take derivitive
y'=1-2x
at x=1
y'=1-2(1)
y'=1-2
y'=-1

the slope is -1
the perpendicular of that slope is what number we can multiply to get -1
-1 times what=-1?
what=1
duh
so
point (1,0) and slope 1
y-0=1(x-1)
y=x-1 is da equation

solve for where y=x-1 and y=x-x² intersect

set equatl to each other since equal y
x-1=x-x²
x²-1=0
factor difference of 2 perfect squares
(x-1)(x+1)=0
set to zero

x-1=0
x=1
we got this point already

x+1=0
x=-1
sub back
y=-1-(-1)²
y=-1-(1)
y=-1-1
y=-2

it intersects at (-1,-2)
Cricetus

The normal line to the parabola [tex]y=x-x^2[/tex] at the point [tex](1,0)[/tex] intersect it second time at the point [tex](-1,-2)[/tex].

The given equation is:

  • [tex]y = x-x^2[/tex]

at point,

  • [tex](1,0)[/tex]

then,

→ [tex]y' = 1-2x[/tex]

So, at (1,0),

→ [tex]y' = 1-2\times 1[/tex]

      [tex]= -1[/tex]

Since,

This is the slope of the tangent, we take its negative reciprocal to get the slope of normal:

= [tex]-\frac{1}{(-1)}[/tex]

= [tex]1[/tex]

The normal line has slope 1 and goes through (1,0):

→ [tex]y-0=1(x-1)[/tex]

→       [tex]y = x-1[/tex]

We want to know where this intersects [tex]y = x-x^2[/tex], we get

→ [tex]x-1=x-x^2[/tex]

→ [tex]x^2=1[/tex]

→ [tex]x = \pm 1[/tex]

hence,

The point corresponding to (1,0) is the one we started with, so we want x=-1:  

→ [tex]x = -1[/tex]

→ [tex]y = x-x^2[/tex]

By substituting the value of "x", we get

→    [tex]= -1-1[/tex]

→    [tex]= -1[/tex]

Thus the answer above is right.

Learn more about Parabola here:

https://brainly.com/question/21685473

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