If x= 0, 1, 2, and 3 and y=1, 1.5, 2, and 2.5, then what is the "y=" function?

Answer:

a) Missing y-values
x = - 2    ==>   y = 5

x = - 1     ==>   y = 2

x =   1    ==>    y = 2

x =   3   ==>    y  = 10

b) Values of x for y = 4:
x = 1.7 and x = -1.7

Step-by-step explanation:

Given the quadratic function,
y = x² + 1

part a)

to find the y-value for a specific x, just plug in that value of x into the right side and simplify

There are four missing values for

The x values and the corresponding y values are
x = - 2    ==>   y =  (-2)² + 1   =  4 + 1 = 5

x = - 1     ==>   y  =  (-1)²  + 1  =  1 + 1 = 2

x =   1    ==>    y  =   (1)²  + 1  = 1 + 1  = 2

x =   3   ==>    y   =  (3)²  + 1  = 9 + 1 = 10

part b)

The graph has each unit on the grid for the x-axis = 0.5
and
each unit on the grid for the  y-axis  = 2

I got this by seeing that between x = 0 and x = 1 there are two units

the value of y at x = -1 or x = 1 is only one vertical unit on the grid and since y = 2 for these values of x, each vertical unit must be 2

Therefore look at the value of y = 4 (which should be two vertical units) on the graph. You can also draw a horizontal line parallel to the x axis at 2 units from the x-axis

It appears that on the positive x-axis for y = 4, x is about halfway between 1.5 and 2 so I would estimate it at 1.75

On the negative x-axis, at y = 4, x is about halfway between -1.5 and -2 so it would be -1.75

One thing to note about quadratic functions is that you get the same value of y for x and -x

So at y = 4, the two possible values of x are
x = 1.75 and x = -1.75

which when using one decimal place will round to 1.7 and 1;75

I am rounding down rather than rounding up since the x value appears be less than 1.75

Mathematically we can solve for x at y = 4 by plugging in values
y = 4 = x² + 1

x² + 1 = 4

x² = 4 - 1 = 3

x =±√3 = ±1.732

which when rounded to 1 decimal place becomes
x = 1.7 and x = -1.7

Answer:

Step-by-step explanation:

a) y = (-2)² + 1 = 5

y = (-1)² + 1 = 2

y = (1)² + 1 = 2

y = (3)² + 1 = 10

b)  4 = x² + 1

x² = 4 - 1 = 3

x = ±√3 = ±1.7