At what point does the normal to y=−4−2x−1x^2 at (1,−7) intersect the parabola a second time?
......The normal line is perpendicular to the tangent line. If two lines are perpendicular their slopes are negative reciprocals -- i.e. if the slope of the first line is m then the slope of the second line is −1/m
......The normal line is perpendicular to the tangent line. If two lines are perpendicular their slopes are negative reciprocals -- i.e. if the slope of the first line is m then the slope of the second line is −1/m
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y = − 4 − 2x − x^2
y ' = − 2 −2x for x = 1 you get
y ' (1) = - 2 - 2*(1) = - 4 then the tangent has slope - 4
Slope of normal is 1/4
Equation of normal is
y + 7 = (1/4)(x - 1) or
y = x/4 - (29/4)
For intersecting point you must solve
y = x/4 - (29/4)
y = − 4 − 2x − x^2
or
x/4 - (29/4) = − 4 − 2x − x^2
x^2 +(9/4)x - (13/4) = 0
4x^2 +9x - 13 = 0
(x-1)(4x + 13) = 0
then x = 1 ( initial point) or
4x+13 = 0
x = -13/4
y = -13/16 - (29/4) = (-13 - 116)/16 = -129/16
y ' = − 2 −2x for x = 1 you get
y ' (1) = - 2 - 2*(1) = - 4 then the tangent has slope - 4
Slope of normal is 1/4
Equation of normal is
y + 7 = (1/4)(x - 1) or
y = x/4 - (29/4)
For intersecting point you must solve
y = x/4 - (29/4)
y = − 4 − 2x − x^2
or
x/4 - (29/4) = − 4 − 2x − x^2
x^2 +(9/4)x - (13/4) = 0
4x^2 +9x - 13 = 0
(x-1)(4x + 13) = 0
then x = 1 ( initial point) or
4x+13 = 0
x = -13/4
y = -13/16 - (29/4) = (-13 - 116)/16 = -129/16
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The parabola is y = -(x^2 + 2x + 4), hence y' = -2(x + 1). The slope at the point (1,-7) is therefore -4, so the equation of the normal line though that point is:
(y + 7) = 1/4 (x - 1)
This will intersect the parabola when
1/4 (x - 1) - 7 = -(x^2 + 2x + 4)
x - 1 - 28 = -4x^2 - 8x - 16
4x^2 + 9x - 13 = 0
(x - 1)(4x + 13) = 0
so the X coordinates of intersection are x=1 (which we already knew), and x = -13/4. The Y value at that second point is -( (-13/4)^2 + 2(-13/4) + 4) = -8 1/16
(y + 7) = 1/4 (x - 1)
This will intersect the parabola when
1/4 (x - 1) - 7 = -(x^2 + 2x + 4)
x - 1 - 28 = -4x^2 - 8x - 16
4x^2 + 9x - 13 = 0
(x - 1)(4x + 13) = 0
so the X coordinates of intersection are x=1 (which we already knew), and x = -13/4. The Y value at that second point is -( (-13/4)^2 + 2(-13/4) + 4) = -8 1/16
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y' = –2 – 2x
y'(1) = –4, so the slpe of the normal is 1/4 and it has the following equation
y = 1/4x + b
–7 = 1/4 + b
b = –29/4
The normal line has equation y = 1/4x – 29/4
x² + 2x + 4 = –1/4x + 29/4
4x² + 8x + 16 = –x + 29
4x² + 9x – 13 = 0. We know x = 1 is a root, so we can find the other by synthetic division
1 |......4......9......–13
..................4.......13
________________
.........4.....13.......|0
4x + 13 = 0, x = –13/4
(–13/4, –129/16)
y'(1) = –4, so the slpe of the normal is 1/4 and it has the following equation
y = 1/4x + b
–7 = 1/4 + b
b = –29/4
The normal line has equation y = 1/4x – 29/4
x² + 2x + 4 = –1/4x + 29/4
4x² + 8x + 16 = –x + 29
4x² + 9x – 13 = 0. We know x = 1 is a root, so we can find the other by synthetic division
1 |......4......9......–13
..................4.......13
________________
.........4.....13.......|0
4x + 13 = 0, x = –13/4
(–13/4, –129/16)
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y=−4−2x−1x^2
dy/dx = -2 - 2*x
At x = 1, dy/dx = -2..the slope of the tangent
Slope of normal = 1/2
Normal is y =(1/2)*x + c
What's c? Normal touches curve at 1,-7
-7 = (1/2) + c
c = -15/2
normal is y = (1/2)*x - (15/2)
Where does -4 - 2*x - x^2 = (1/2)*x - (15/2)?
(7/2) - (5/2)*x - x^2 = 0
x^2 + (5/2)*x - (7/2) = 0
2*x^2 + 5*x - 7 = 0
(x - 1)*(2*x + 7) = 0
x = 1 (we knew that) or x = -7/2
Normal is y = (1/2)*x - (15/2)
At x = -7/2
y = -(7/4) - (30/4) = - 37/4
Thus the point of intersection is ( -1/2, -37/4) <<
dy/dx = -2 - 2*x
At x = 1, dy/dx = -2..the slope of the tangent
Slope of normal = 1/2
Normal is y =(1/2)*x + c
What's c? Normal touches curve at 1,-7
-7 = (1/2) + c
c = -15/2
normal is y = (1/2)*x - (15/2)
Where does -4 - 2*x - x^2 = (1/2)*x - (15/2)?
(7/2) - (5/2)*x - x^2 = 0
x^2 + (5/2)*x - (7/2) = 0
2*x^2 + 5*x - 7 = 0
(x - 1)*(2*x + 7) = 0
x = 1 (we knew that) or x = -7/2
Normal is y = (1/2)*x - (15/2)
At x = -7/2
y = -(7/4) - (30/4) = - 37/4
Thus the point of intersection is ( -1/2, -37/4) <<
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it happens