#### Question

Find the minimum and maximum values of the quadratic function

y = x^2-4x+3 (0\leq x \leq 5 )

Question

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