Well the y-intercept is +2
And the slope of the blue line is 1/2
So y=1/2x+2
Slope is rise/run (up then ((right)))
And the intercept is found where the line intercects the y line.
There are two ways you can write it:
1. 30 + 6 + (1 × 1/10) + (3 × 1/100)
2. 30 + 6 + 0.1 + 0.03
Either is correct :)
If you begin with the basic equation of a vertical parabola: y-k=a(x-h)^2, where (h,k) is the vertex, then that equation, when the vertex is (-3,-2), is
y + 2 = a (x + 3)^2. If we solve this for y, we get
y = a(x+3)^2 - 2. Thus, eliminate answers A and D. That leaves B, since B correctly shows (x+3)^2.
One meaning of a 'linear' equation is that if you draw the graph
of the equation, the graph will be a straight line.
That's an easy way to test the equation . . . find 3 points on the
graph, and see whether they're all in a straight line.
This equation is y = 4 / x .
To find a point on the graph, just pick any number for 'x',
and figure out the value of 'y' that goes with it.
Do that 3 times, and you've got 3 points on the graph.
Here ... I'll do 3 quick points:
Point-A: x = 1 y = 4 / 1 = 4
Point-B: x = 2 y = 4 / 2 = 2
Point-C: x = 4 y = 4 / 4 = 1
Look at this:
Slope of the line from point-A to point-B
= (change in 'y') / (change in 'x') = -2 .
Slope of the line from point-B to point-C
= (change in 'y') / (change in 'x') = -1/2 .
The two pieces of line from A-B and from B-C don't even have
the same slope, so they're not pieces of the same straight line !
So my points A, B, and C are NOT in a straight line.
So the equation is NOT linear.
Try it again with three points of your own.
