Draw a linear function where -2 ≤ x ≤ 4 and 0 ≤ y ≤ 3. At which point does the line intersect the y-axis? Two cases!

Question

Draw a linear function where -2 ≤ x ≤ 4 and 0 ≤ y ≤ 3. At which point does the line intersect the y-axis? Two cases!

Answer 1

From my understanding, you want the lines to exist between the given boundaries.

So first, you would draw the boundary lines defined above. The boundaries result in a rectangle with the corners as (-2,0), (-2,3), (4,3) and (4,0).

The linear equation is defined by y = mx+b where m is the slope and b is the y intercept.

The slope(m) of can be determined from rise/run or (y₂-y₁) / (x₂-x₁).

If the diagonal is from (-2,0) to (4,3) then we can define them as (-2,0) = (x₁,y₁) and (4,3) = (x₂,y₂).

The slop becomes (3-0)/(4-(-2)) = 3/6 = 1/2

b can be determined by plugging in a known coordinate that the line passes through (one of the 2 corners we know so here I will choose (-2,0) for simplicity) and the known m value then solving for b.

y = mx+b

0 = (1/2)(-2) + b

0 = -1+b

b = 1

So, the first equation is y = (1/2)x+1

The second equation is the other diagonal. We repeat getting m and b as before but for the other two corners ((-2,3) and (4,0)).

We can define them as (-2,3) = (x₁,y₁) and (4,0) = (x₂,y₂).

The slop becomes (0-3)/(4-(-2)) = -3/6 = -1/2

b can be determined by plugging in a known coordinate that the line passes through (one of the 2 corners we know so here I will choose (4,0) for simplicity) and the known m value then solving for b.

y = mx+b

0 = (-1/2)(4) + b

0 = -2+b

b = 2

So, the second equation is y = (-1/2)x+2

I hope this answers your question. Let me know if you need any more help.

Answer 2

The short answer is that the any given line could intersect the y axis between the points 0 and 3.

Here are two examples. (Y=MX+B)

Y=2x+3, This will intercept the y axis at point (0,3)

Y=9x+0, This will intercept the y axis at point (0,0)

the equation for the line can be just about anything so long as the B is between 0 and 3.

Now I don't exactly know if this will answer your question as the equations you have provided would create a box of possible points. The four corners of the box are :(-2,0) (-2,3) (4,3) (4,0). However each individual equation does not create a line itself. Rather they boy create area that would render the equation true, but due to lacking an X or Y they effectively go on forever within a confined area. For example -2 ≤ x ≤ 4 would be and point between (-2,y) and (4,y), with Y having an infinite number of values. I hope this answers your question.