 Hello friends and how are you all doing today? The question says, find the equation of the tangent and nominal to the given curves at the indicated points. Now here we are given the equation of the curve as x raised to the power 4 minus 6xq plus 13x square minus 10x plus 5 at 1 comma 3. Let's proceed with the solution quickly. Let us rewrite y is equal to x raised to the power 4 minus 6xq plus 13x square minus 10x plus 5. Now let us differentiate y with respect to x. And on doing so we get dy by dx is equal to 4xq minus 18x square plus 26x minus 10, isn't it? Now we need to find out the value of dy by dx at point 1 comma 3. It is dy by dx at since we only have x in R, dy by dx the value will be 4 1q minus 181 square plus 26 into 1 minus 10 which is 4 minus 18 plus 26 minus 10. On solving we have 30 minus 28 that is equal to 2. So that means slope of tangent is 2. So slope of normal will be minus 1 upon dy by dx that is minus 1 upon 2. Now since we have slope of both tangent and normal we can find the equation of both of them. So we have equation of tangent as y minus y1 which is 3 equal to m that is slope x minus x1 that is 1. Similarly we will be solving it. So we have y minus 3 equal to 2x minus 1 which gives us y is equal to 2x minus 2 plus 3 will give us plus 1. In the same manner we will be finding out equation normal is y minus y1 equal to slope of the normal x minus x1 which gives us 2y minus 6 is equal to minus x plus 1 which further gives us the value as x plus 2y minus 6 here it will become minus 1 will be minus 7 equal to 0. So the answer for the second part is equation for tangent is y is equal to 2x plus 1 and for normal it is x plus 2y minus 7 is equal to 0. So this completes our solution hope you understood the whole concept well do take care of your calculations and have a very nice day ahead.