For a graph C with equation y=3/(5-3x)^2, find the the equation of the line normal to the graph at point P, where x=2. Give your answer in the form ax+by+c=0
Find out the value of the y coordinate where x=2y=3/(5-3(2))^2=3Differentiate function in order to find the gradient of the graph at point Py=3/(5-3x)^-2Use chain rule where u=5-3xdu/dx=-3dy/dx=3u^-2Hence dy/dx=-6u^-3*-3dy/dx=18(5-3x)^-3dy/dx at x=2 is 18(5-3*2)^-3=-18Take negative reciprocal of 18 to find gradient of line normal to graph-18-->1/18y-y1=m(x-x1)Substitute in valuesy-3=1/18(x-2)18y-54=x-218y-x-52=0-x+18y-52=0
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All answers ▸The point P (4, –1) lies on the curve C with equation y = f( x ), x > 0, and f '(x) =x/2 - 6/√x + 3. Find the equation of the tangent to C at the point P , giving your answer in the form y = mx + c. Find f(x)
Susan is researching the population growth of a city. She proposes that x, the number of people in the city, t years after 2017 is given by x=250,000e^(0.012t) A.population in 2017 B.population in 2020 C.During which year would the population have doubled
Integrate cos^2(x)
Derive the following with respect to x1: y=(x1*x2)/(x1+x2).