00:01
So we are given with function y equals to f of x, which is equals to 5x square over x squared plus 3, and we need to find x and y intercept with the function.
00:08
So x intercept will occur when y is 0.
00:10
So if you substitute y is 0, we'll have 5x square over x square plus 3.
00:16
And this will occur 0 when this numerator is 0, and therefore we get 5x square as 0 and that gives x as 0.
00:24
So this is our x intercept.
00:26
And now to find the y intercept, we have to put x as 0.
00:30
If you substitute x as 0, we already know that is y 0 and only intercept x and y axis it at 0 .0.
00:37
And since it is intersecting at 0 .0, it says that the curve touches at 0 .0.
00:43
So, touches at 0 .0 at this point.
00:47
Okay.
00:47
Now the type of symmetry.
00:49
So the type of symmetry to find the symmetry about the y -axis, we can substitute x as minus x or we can replace x with minus x.
00:56
And if you replace it we get 5 minus x square will again be x squared divided by minus x square again x squared plus 3 and this says that this is equals to f of x and from this we can say the given function is symmetric about y axis since we get the same result either we substitute minus x or we substitute x and we are getting the same y so it is symmetric about y it is not symmetric about x -axis since there is no symmetricity in y okay now next is vertical and horizontal asymptote.
01:25
To find the vertical asymptote, we have to check where the function is not defined or the denominator is 0.
01:31
So, denominator is x square plus 3 and if we equate with 0, we get x square as minus 3 and this is not possible, hence there is no vertical asymptote.
01:40
To find the horizontal asymptote, we see here degree of numerator is, degree of numerator is equal to degree of denominator since both have x squaredum.
01:50
The asymptote is obtained, that is x is equal, sorry, y is equals to that will give us horizontal asymptote is the highest degree coefficient.
01:58
So we need to write the coefficient of let's say numerator or coefficient of denominator.
02:03
So what is coefficient of numerator of highest degree? it is 5 and coefficient of numerator of highest degree is 1.
02:08
So we get 5.
02:10
So this is our horizontal asymptote.
02:12
Now next move to compute the first derivative and use your answer to determine where the function is increasing, decreasing.
02:18
So your answer in a work line.
02:19
Okay.
02:20
So we have function as 5x square over.
02:24
X square plus 3.
02:26
Now differentiating this.
02:27
So this is u over v so we'll apply u or v so we have to take u out sorry v out differentiation of u that is 5 times 2x right x square differentiation minus then we'll take 5x 2x 2x divided by v square so x square plus 3 whole square.
02:45
So this is our f -dazof x now if you simplify this so we can write this as on multiplying this inside this is 10x times x square so that gives us 10x cube then 3 times 10 x is 30 x and this is minus 10x cube divided by x square plus 3 whole square so we can cancel these two terms now to check the function the point where it is increasing or decreasing so let me write our f dash of x is 30x over x square plus 3 whole square now here with this x square plus 3 whole square term is always positive so we can neglect it so we'll be left with only 30x and now this will be negative when x is less than 0 and positive when x is greater than 0.
03:28
So, function is increasing.
03:29
So if you draw a number line, let's say this is 0 since we have the point 0 and rest all the points are here similarly in this direction.
03:38
So if x is less than 0, we have function negative and hence it is decreasing in this region.
03:44
So let's say decreasing here.
03:46
And when x is greater than 0, we have positive value of dash of x.
03:49
So function is increasing in this region.
03:52
Okay.
03:53
Now let's move to the third one, which is asking us to compute second order derivative and use your answer to determine where the function is concave up or concave down.
04:01
So we have already calculated the first order.
04:03
Now let's calculate the second order.
04:05
So again we have to apply u over v...