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)
<h3>
Answer: x-2y = -8</h3>
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Explanation:
Multiply both sides by 2 to clear out the fraction
y = (1/2)x+4
2y = 2[ (1/2)x + 4 ]
2y = 2*(1/2)x + 2*4
2y = x + 8
Then move the x term over to the left side
2y = x+8
2y-x = 8
-x+2y = 8
Optionally we can multiply both sides by -1
-x+2y = 8
-1*(-x+2y) = -1*8
x-2y = -8
This is in standard form Ax+By = C with A = 1, B = -2, C = -8
The reason why I multiplied both sides by -1 was to make A > 0 which is what some textbooks use as convention. Of course -x+2y = 8 is equally valid too.
Your answer is 98. Is there a multiple choice?
-2.8 • -2.8 • -2.8 • -2.8 • -2.8 • -2.8
=481.890
<span>3.1415926535897932384626</span>
