Maxima and Minima of Quadratic Function

Question

When 0≤x≤5, find the corresponding set of y values (the range)

1. Given y=x^2-6x+4

Please help my worksheet is very confusing. Please explain thoroughly also 🙏

Answer 1

y= x^2 -6x+4 with x from 0 to 5

it's an upward opening parabola

y= x^2 -6x + 9 +4-9

= (x-3)^2 -5 with vertex = minimum point = (3,-5)

range of y is from -5 to 4 for when x is between 0 and 5

y=-5 is the global and local minimum, when x=3

for x=0, y=(0-3)^2-5 = 9-5=4

for x=5, y = (5-3)^2-5 = 4-5=-1

Answer 2

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