Equation of graph is - y=x^3+x^2-5x-2 1. Show that there is a turning point when x=1 and find the x co-ordinate of the turning point 2.Find the equation of the tangent at the point (0,-2) 3.Find the equation of the normal at the point (2,0) Thanks
Equation of graph is - y=x^3+x^2-5x-2 1. Show that there is a turning point when x=1 and find the x co-ordinate of the turning point 2.Find the equation of the tangent at the point (0,-2) 3.Find the equation of the normal at the point (2,0) Thanks
1. Differentiate the function to get 3x^2 + 2x -5, then equate it to zero and solve teh quadratic , which should give you 2 turning points , one being x =1.
2. sub in x= 0 into 3x^2 +2x- 5 , you get -5 , so the gradient of the tangent is -5, therefore y= -5x + c, sub in your coordiante (0, -2) to find C and that is your equation.
3. the gradient of the normal is the negative reciprocal of the gradient of the tangent , which is 1/5 so y= 1/5x+c , sub in (2,0) to find your equation.
1. Differentiate the function to get 3x^2 + 2x -5, then equate it to zero and solve teh quadratic , which should give you 2 turning points , one being x =1.
2. sub in x= 0 into 3x^2 +2x- 5 , you get -5 , so the gradient of the tangent is -5, therefore y= -5x + c, sub in your coordiante (0, -2) to find C and that is your equation.
3. the gradient of the normal is the negative reciprocal of the gradient of the tangent , which is 1/5 so y= 1/5x+c , sub in (2,0) to find your equation.