 Hello and welcome to the session. In this session we will discuss the following question that says sketch the graph of the function y equal to square root of 16 minus x square. Let's start with the solution now. The function given to us is y equal to square root of 16 minus x square and we have to sketch the graph for this function. Let this be equation one. For the sketching of the graph for this function we would follow certain steps. In the first step we check the symmetry of a given function. As you can see that equation one contains only even powers of x so the curve is symmetrical about the y axis. In the next step we check whether the curve passes through the origin or not. So for this we put x equal to 0 and y equal to 0 in equation one and we get 0 is equal to square root of 16 minus 0 that is we have 0 is equal to 4 which is obviously not true therefore we can say that the point 00 does not satisfy equation one the curve does not pass through the origin. In the next step we find the points of intersection of the curve with the now put in 0 in equation one that is this equation we get y is equal to square root of 16 minus 0 which is equal to square root of 16 that is plus minus 4. So we can say that the curve that's the y axis at points 0 4 0 minus 4 then next putting y equal to 0 in equation one we get 0 is equal to square root of 16 minus x square that is 16 minus x square is equal to 0 or we can say we have x square equal to 16 which gives us the values of x as plus minus 4. So we can say that the curve that's the x axis 4 0 and minus 4 0 now next in which the curve is increasing decreasing now as we had y equal to square root of 16 minus x square from here we have dy by dx is equal to 1 upon 2 into square root of 16 minus x square into minus 2x this 2 into cancels so we have dy by dx is equal to minus x upon square root of 16 minus x square now for finding the intervals in which the curve is increasing we take dy by dx greater than 0 that is minus x upon square root of 16 minus x square is greater than 0 we can say that minus x is greater than 0 which means that x is less than 0 now for finding the intervals in which the curve is decreasing we have dy by dx less than 0 that is minus x upon square root of 16 minus x square is less than 0 which means that minus x is less than 0 all you can say that x is greater than 0 so we can say the curve is increasing when we have x less than equal to 0 it is decreasing when we have x greater than equal to 0 now in the next step dy by dx is equal to minus x upon square root of 16 minus x square now from here we have d2y by dx2 is equal to minus x into minus 1 by 2 into 16 minus x square the whole to the power of minus 3 by 2 into minus 2x the whole plus 16 minus x square to the power of minus 1 by 2 into minus 1 so this would further give us d2y by dx2 as minus 16 upon 16 minus x square to the power of 3 by 2 now when we have dy by dx equal to 0 this means minus x upon square root of 16 minus x square is equal to 0 that is we get x equal to 0 now d2y by dx2 equal to 0 would be less than 0 that is when we put x equal to 0 here we will get a negative value for this so this means that y is maximum when x is equal to 0 so we can say that x equal to 0 is a maximum point the maximum value of y would be equal to square root of 16 minus 0 that is equal to 4 that we can say 0 4 is a maximum point now again we have y is equal to square root of 16 minus x square now y is defined when we have 16 minus x square greater than equal to 0 that is when we have x square less than equal to 16 or we can say that modulus of x is less than equal to 4 which means that x lies between minus 4 and 4 that is x is greater than equal to minus 4 and less than equal to 4 that is no portion of the curve lies on the left of x equal to minus 4 or on the right of x equal to 4 now in the next step we prepare a table of values to find a few points we had y equal to square root of 16 minus x square as the function for x equal to minus 4 in this function we have the value of y as 0 for x equal to minus 3 value of y is 2.6 for x equal to minus 2 the value of y is 3.5 for x equal to minus 1 the value of y that we obtain is 3.9 for x equal to 0 the value of y so obtained is 4 then for x equal to 1 the value of y is 3.9 for x equal to 2 the value of y is 3.5 for x equal to 3, the value of y is 2.6. For x equal to 4, the value of y is 0. Now we will plot these points. So first we plot the point which coordinates minus 4, 0. So this is the point minus 4, 0. Next point that we have is minus 3, 2.6. This point represents the point which coordinates minus 3, 2.6. Now the next point that we have is minus 2, 3.5. Now let us plot this point. The point minus 2, 3.5 is this point. Then we have minus 1, 3.9. This point is minus 1, 3.9. Now next we have the point 0, 4. This is the point 0, 4. Next is 1, 3.9. So this point is 1, 3.9. Then we have a point which coordinates 2, 3.5. So this point represents the point which coordinates 2, 3.5. Next point to be plotted is 0.6. So is 3, 2.6. Then we have the point 4, 0. So this is the point 4, 0. Now we will join these points by a free hand curve. So we have got this curve of symmetrical about the y-axis. Then it does not pass through the origin. It intersects the x-axis at the points 4, 0 and minus 4, 0. And the y-axis at the point 0, 4. And we observe that this curve is increasing when x is less than equal to 0 and decreasing when x is greater than equal to 0. This point 0, 4 is the maximum point and we also observe that no portion of the curve lies on the left of x equal to minus 4 that is on this side and no portion of the curve lies on the right of x equal to 4 that is on this side. Now the given function was y equal to square root of 16 minus x square. Now when we square both sides we get y square equal to 16 minus x square. Further we get x square plus y square is equal to 16. This equation represents a circle with centre 0, 0 and radius 4. And graph of this function as we have drawn is the upper half of this circle that is x square plus y square equal to 16. So this completes the sketching of the graph. Hope you have understood the solution of this question.