Question
Find the minimum and maximum values of the quadratic function
y = x^2-4x+3 (0\leq x \leq 5 )
Find the minimum and maximum values of the quadratic function
y = x^2-4x+3 (0\leq x \leq 5 )
y = x^2-4x+3
=(x-2)^2-1
So the parabola opens upwards and
the axis of symmetry is x = 2 <\dfrac{5}{2} = 2.5, which is less than the midpoint of the domain 0\leq x \leq 5
Then,
y_{min} = -1, when x = 2
y_{max} = (5-2)^2-1= 8, when x = 5