Find the turning points of the curve (x^3)/3 + x^2 -8x + 5

Step one, we need to assess what the question is asking us to do, in this case, find the turning points.The turning points on the graph are where the gradient is equal to zero, so in order to find the turning points, we need to se the gradient of the graph to zero.Step 2 therefore is to find the gradient, we do this by differentiation. Remember, power down in front, one off the powerd/dx x³/3 + x² - 8x + 5 = x² + 2x -8Step 3 is to set the gradient we have found equal to zero and solve for xx² + 2x -8 = 0in this case, we can factorise, what could you use if you couldn't factorise? (quadratic formula)(x - 2)(x + 4) = 0therefore, x = 2 or x = -4 have we answered the question? Not yet, we need to find the corresponding y values to our x values. Step 4, We plug our x - values back into our equationfor x = 2,y = 2³/3 + 2² - 82 + 5 = -4.33for x = -4y = (-4)³/3 +(-4)² - 8(-4) + 5 = 31.67
Finally, state the answer:turning points: (2, -4.33) and (-4, 31.67)