How many real number solutions does the equation y=2x^2−8x+8 have?

1 real solution

Step-by-step explanation:

y=2x^2−8x+8

We can use the discriminant to determine the number of real solutions

b^2 -4ac

a =2  b = -8  c=8

(-8)^2 - 4(2)(8)

64 - 64

0

Since the discriminant is 0 there is 1 real solution

>0 there are 2 real solutions

< 0 two complex solutions

x -

median = 1.25

range = 12-14.5

interquartile = q12-q14.5

y -

median = 1

range = 15-17

interquartile = q15-q17