Find the coordinates of the minimum point of the curve y = 3x^(2) + 9x + 10

(Note: in this answer I will use the notation y' to represent the differential of y, and by the same reasoning y'' denotes the differential of y', or the second derivative of y.)

A curve's maximum or minimum will be found at the stationary point of the curve (for a continuous function), where the gradient of the curve is flat and equal to zero.

We can find the gradient of a curve by differentiation, where the first derivative is found by the rule of 'bring the power down and minus 1 from the power.' Formally, this can be written as if y = Axⁿ, then the differential of y with respect to x is y' = nAx^n-1. Consider the curve y = 3x² + 9x + 10 to see how this works in practice, first including the equation's hidden powers for more clarity:

y = 3x² + 9x¹ + 10x⁰

y' = (2)(3)x^2-1 + (1)(9)x^1-1 + (0)(10)x^0-1

= y' = 6x + 9, as x⁰ = 1, and the last term disappears as it is multiplied by zero, constants (terms with no 'x' included) will always drop out once differentiated.

As stated earlier, the maximum or minimum is found where the curve is stationary, and the gradient is equal to zero, and so we set our newly derived gradient equal to zero, and rearrange the equation to find the value of x at this point:

6x + 9 = 0 (minus 9 from both sides)

6x = -9 (divide both sides by 6)

x = -9/6 = -3/2, and so the curve is stationary at the point where x is = to -3/2. What is the value of y at this point? Simply substitute this newly found x value into the original equation of the curve:

y = 3(-3/2)² + 9(-3/2) + 10 = (27/4) - (27/2) + 10 = 13/4. So there is a stationary point at (-3/2, 13/4).

Finally, in order to ensure that this point is a minimum point, we need to perform the second derivative test. This test states that we must differentiate y again (differentiate y'), if:

y'' > 0, then the curve is convex and minimised at that point.

y'' < 0, then the curve is concave and maximised at that point.

Therefore, lets differentiate y' = 6x + 9 again:

y'' = 6 which is > 0 and so the curve is in fact minimised, thus we can conclude that the minimum point of the curve is found at the point (-3/2, 13/4)