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
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