Maximum and minimum values questions with answers

Questions

1. Question 1Read the turning point

   Write down the coordinates of the maximum point of the curve $y = -3(x - 2)^2 + 5$.

   Show the worked solution

   1. Step 1: The squared bracket is never negative, and it is 0 when the bracket is 0$x - 2 = 0$ when $x = 2$
   2. Step 2: A square taken away: y is largest when the square is 0$y = 5$The maximum point is $(2, 5)$.
2. Question 2Minimum value: complete the square

   By completing the square, find the minimum value of $y = x^2 - 4x - 5$.

   Show the worked solution

   1. Step 1: Complete the square: halve the coefficient of x$x^2 - 4x - 5 = (x - 2)^2 - 4 - 5$$= (x - 2)^2 - 9$
   2. Step 2: The bracket squared is never negative: the minimum is when it is 0$(x - 2)^2 \geq 0$$\text{Minimum value} = -9$
3. Question 3Maximum value: take out −1 first

   By completing the square, find the maximum value of $y = 3 + 4x - x^2$.

   Show the worked solution

   1. Step 1: Take out −1, then complete the square inside$3 + 4x - x^2 = -(x^2 - 4x) + 3$$= -[(x - 2)^2 - 4] + 3$$= -(x - 2)^2 + 4 + 3$$= -(x - 2)^2 + 7$
   2. Step 2: The bracket squared is never negative: the maximum is when it is 0$(x - 2)^2 \geq 0$$\text{Maximum value} = 7$
4. Question 4Where it happens

   By completing the square, find the value of $x$ at which $y = x^2 + 10x + 12$ has its minimum value.

   Show the worked solution

   1. Step 1: Complete the square: halve the coefficient of x$x^2 + 10x + 12 = (x + 5)^2 - 25 + 12$$= (x + 5)^2 - 13$
   2. Step 2: The bracket squared is never negative: the minimum is where the bracket is 0$(x + 5)^2 \geq 0$$x + 5 = 0$$x = -5$