This is an ADDMATHS QUESTION so I'm not sure if anyone can help me but please try to ㅠ ㅠ: Find the equation of the normal to the curve y=ln(2x^2 - 7) at the point where the curve crosses the positive x-axis. Give your answer in the form ax+by+c=0 , where a, b and c are integers.

You can go through the effort of determining the zero of the function analytically and evaluating an analytic expression for the derivative at that point, or you can let a graphing calculator do that heavy lifting. Since the numbers have to be "nice" for your equation to have the desired form, it is easy to know what to round to in the event that is necessary (it isn't).

We find the positive zero-crossing at x=2, and the slope of the curve at that point to be 8. Thus the line will have slope -1/8 and can be written as
.. x +8y -2 = 0

TheAakash TheAakash · Answer 2 · 2018-03-02T08:08:46+01:00

This equation crosses the positive x-axis at (0, 2). I used a graphing calculator, you could brute force it if you'd like.

First, take the derivative of the equation given:
[tex]y = ln(2x^2-7)[/tex]
[tex]y' = \frac{4x}{2x^2-7} [/tex]

At 2, the function has a derivative of 8 (just plug in 2 to y'). So, the perpendicular to that has to have the slope -1/8 because of the negative-inverse rule. We have this until now:
[tex]y = -\frac{1}{8}x -2[/tex]

Now, we have to get a 0 on the right side, so, just get everything on the other side, and you are done.