Here is what I found, look at the end to see what my question is:(adsbygoogle = window.adsbygoogle || []).push({});

At what point does the normal line to the graph of y= -4 - 3x + 4x^2 at (1, -3 ) intersect the parabola a second time?

So far I have tried to find the derivative of the parabola which is:

8x-3 (this corresponds to the slope of the tangent line). Therefore now I replaced x by 1 because of the point I am given. This gives:

8(1)-3 = 5

So that is the slope of the tangent line. Now to know the equation of this line I do:

y=5x+b , replace x and y by the point I am given and solve for b.

-3=5(1)+b

b=-8

Thus, the equation is: y=5x-8

The normal is perpendicular to the tangent at (1,-3) so I take the negative reciprocal of the slope to find the slope of the normal:

5 becomes -1/5 (reciprocal). And now I get y=-1/5+b for the normal. Replace x and y by the point (1,-3):

y=-1/5x+b

-3=-1/5x+b

-3=-1/5(1)+b

-14/5=b

The equation for the normal is: y=-1/5x-14/5

Now I want to find when this line and the parabola meet so:

-1/5x-14/5 = -4 - 3x + 4x^2

0 = -6/5 -14/5x + 4x^2

Using (-b+/-sqrt(b^2-4ac))/2a

x1 = 1

x2 = -3/10

x1 = 1 I already know because that is the point where the normal, the tangent, and the parabola intersect. And x2 = -3/10 is the other point where only the parabola and the normal intersect.

To find the y-coordinate of -3/10 I just insert it into the equation and get:

y=-1/5(-3/10)-14/5

y= -137/50

Thus the other point is: (-3/10,-137/50)

I would just like to know if the above calculations are correct and that I am doing the proper steps.

Thanks a lot,

Cedrick O'Shaughnessy

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# Homework Help: Normal Line and a Function

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