Answer:
Two of the statements apply:
- Second statement: Binh should have graphed the y-intercept of y = x - 3 at (0, - 3)
- Last statement: Binh should have found the point of intersection to be (1, - 2).
Explanation:
1. Line 1
[tex]y=-\dfrac{1}{2}x-\dfrac{3}{2}[/tex]
2. Line 2
[tex]y=x-3[/tex]
3. Binh’s Graph on a coordinate plane, a line with equation y = x - 3 goes through (- 3, 0) and (0, 3).
This is wrong because the point (-3, 0) is not on the line y = x - 3:
- x = -3 ⇒ y = -3 - 3 = -6 ⇒ (-3,-6)
The other point, (0, 3) is wrong too
- x = 0 ⇒ y = 0 - 3 = - 3 ⇒ (0, -3) This is the y-intercept
4. A line with equation y = -(1/2)x - (3/2) goes through (- 3, 0) and (1, -2).
Check the points:
- x = -3 ⇒ y = -(1/2)(-3) - 3/2 = 3/2 -3/2 = 0 ⇒(-3,0) [tex]\checkmark[/tex]
- x = 1 ⇒ y = -(1/2)(1) - 3/2 = -1/2 -3/2 = -2 ⇒ (1, - 2)[tex]\checkmark[/tex]
Both points are correct
5. Binh says the point of intersection is (0, –3). Which statements identify the errors Binh made?
Binh listed the coordinates in the wrong order when describing the point of intersection on his graph?
- No, his error was on the procedure, look below
Binh should have graphed the y-intercept of y = x - 3 at (0, - 3)?
Correct!
- This was one of the errors, as shown above the y-intercept of the line y = x - 3 is (0, -3) and not (0, 3).
Binh should have graphed the y-intercept of y = x - 3 at (0, 1)?
- No. it is (0, 3) as stated above.
Binh should have graphed the y-intercept of y = -(1/2)x -3/2 at (0, -1/2)?
No, this is wrong:
- The y-intercept is y = 0 - 3/2 = -3/2 ⇒ (0, -3/2)
Binh should have found the point of intersection to be (1, - 2).
Correct!
Yes, the solution of the equation is (1, -2) and he should have found that intersection point. Proof:
Thus, the point of intersection is (1, -2).
