y7xx 1 a2 find the equation of the line tangent to the given curve at xa use a graphing utility

Step 1: Find the derivative of the function y = 7x/(x-1).

Y7xx 1 a2 find the equation of the line tangent to the given...

Transcript

00:01 We have been given a function y equal to 7x by x minus 1 and we have to find the tangent line of this curve at the point x equal to 2 to find the tangent line we have to first find the slope of the tangent line and to find the slope of the tangent line we have to differentiate this function y with respect to x therefore ddx of 7x by x minus 1 is given by we will consider the numerator at the first function and the denominator at the first function and the denominator as the second function so the derivative of the first function is 7 times the second function minus first function and derivative of the second function is 1 by the square of the second function in the denominator the value of d .dx at x equal to 2 will be we will substitute 2 4 x so this is equal to 7 minus 14 which is equal to minus 7 therefore the slope of the tangent line is given by minus 7 and if the if a function passes through the point x1 y1 and has slope m then the equation of the line is given by y minus y1 is equal to m times x minus x1 here x1 is 2 and y1 is the value of y at 2 at x equal to 2 which is given by 4 so the x1y 1 is here 2 and 14 the tangent line passes through this point 2 14 and has a slope minus 7 therefore the equation of the tangent line is given by and x equal to 2 is given by y minus 14 is equal to minus 7 times x minus 2 which is equal to 1 y plus 7 x and 14 that is equal to y plus 7 x that is equal to y plus plus 7x is equal to 28 now we have to draw the graph of the function y equal to 7x by x minus 1 to find the graph we will find a few points that satisfies this curve we will find assume some values of x and we will find the corresponding value of y at those points suppose we are taking minus 1 the value of y at minus 1 is given by minus 7 by minus 2 that is equal to 7 by 2 or 3 .5 at 0 the functional value is 0 at 1 the function is 1 minus 1 this is undefined actually if we equate the denominator to 0 this gives the vertical asymptote of the function so x equal to 1 line is the vertical asymptote of this function at x equal to to 2 the function is given by 14 at x equal to 3 the functional value is given by 21 divided by 2 that is 10 .5 at 4 the functional value is given by 28 divided by 3 that is equal to 9 .333 so the pair of points that satisfies the function the curve of the function are minus 1 3 .5 02014 and 310 5 49 .33 and the vertical estimate of the function is x equal to 1 we will use these points to draw the graph of the function and to plot the tangent line we will find the intercept of this line and the intercepts are given by we will write this equation in decept form an intercept from this equation can be written as x by 4 plus y by 28 is equal to 1 so the intercept of this line are the x intercepts and y intercepts are 4 0 and 028 we will use this points to draw the tangent line at the point 240 consider this as the horizontal axis x and this as y and consider this is 5, 10, 15, 20, 25 and here 28.

05:56 And along this line, minus 1, 1, 2, 3, 4, 5.

06:07 We'll plot these points as obtained.

06:10 Minus 1, 3 .5 will lie somewhere here.

06:15 0 is this point and 214 will be.

06:23 Somewhere here 310 .5 and 4 9 .3 3 .3 and x equal to 1 and x equal to 1 here will be a and let us consider this horizontal vertical line at x equal to 1 is the vertical asymptote of this function so the function is given by so the curve of the function will be of this form going up to minus infinity as this this is the vertical asymptote and this is coming from minus up to infinity like this...

Question

y = 7x/(x-1); a = 2 Find the equation of the line tangent to the given curve at x = a. Use a graphing utility to graph the curve and the tangent line on the same set of axes.

Question

y = 7x/(x-1); a = 2 Find the equation of the line tangent to the given curve at x = a. Use a graphing utility to graph the curve and the tangent line on the same set of axes.

Please give Ace some feedback