grade 9 mathematics unit 4 - linear relations\n8. match each equation with the appropriate graph.\n(i) $y =…

grade 9 mathematics unit 4 - linear relations\n8. match each equation with the appropriate graph.\n(i) $y = 2x + 1$ (ii) $y = -2x + 1$ (iii) $y = 2x - 1$\n(iv) $y = -2x - 1$ (v) $y = 2x$ (vi) $y = -2x$\n(vii) $y = 2$ (viii) $y = 1$ (ix) $y = -1$\nfirst row of graphs\nequation: ____ equation: marked ix equation: ____\nsecond row of graphs\nequation: ____ equation: ____ equation: marked vii\nthird row of graphs\nequation: ____ equation: marked viii equation: ____
Answer
Explanation:
To solve this, we analyze each linear equation's slope ((m)) and y - intercept ((b)) in the slope - intercept form (y = mx + b) (for horizontal lines (y = k), (m = 0) and (b=k)):
Step 1: Analyze horizontal lines ((y = k))
- For (y = 2) (vii), (y = 1) (viii), (y=-1) (ix): These are horizontal lines (slope (m = 0)) crossing the y - axis at (y = 2), (y = 1), and (y=-1) respectively.
- The graph with the horizontal line at (y = 2) matches (y = 2) (vii). The graph with the horizontal line at (y = 1) matches (y = 1) (viii). The graph with the horizontal line at (y=-1) matches (y=-1) (ix).
Step 2: Analyze lines with positive slope ((m>0))
- Equations (y = 2x+1) (i), (y = 2x - 1) (iii), (y = 2x) (v): All have a slope (m = 2) (positive, steep line).
- (y = 2x) (v) has a y - intercept (b = 0) (passes through the origin ((0,0))).
- (y = 2x+1) (i) has a y - intercept (b = 1) (crosses the y - axis at ((0,1))).
- (y = 2x - 1) (iii) has a y - intercept (b=-1) (crosses the y - axis at ((0, - 1))).
- Find the graphs with a steep positive slope. The graph passing through the origin with a steep positive slope matches (y = 2x) (v). The graph crossing the y - axis at ((0,1)) with a steep positive slope matches (y = 2x + 1) (i). The graph crossing the y - axis at ((0,-1)) with a steep positive slope matches (y = 2x-1) (iii).
Step 3: Analyze lines with negative slope ((m<0))
- Equations (y=-2x + 1) (ii), (y=-2x-1) (iv), (y=-2x) (vi): All have a slope (m=-2) (negative, steep line).
- (y=-2x) (vi) has a y - intercept (b = 0) (passes through the origin ((0,0))).
- (y=-2x + 1) (ii) has a y - intercept (b = 1) (crosses the y - axis at ((0,1))).
- (y=-2x-1) (iv) has a y - intercept (b=-1) (crosses the y - axis at ((0,-1))).
- Find the graphs with a steep negative slope. The graph passing through the origin with a steep negative slope matches (y=-2x) (vi). The graph crossing the y - axis at ((0,1)) with a steep negative slope matches (y=-2x + 1) (ii). The graph crossing the y - axis at ((0,-1)) with a steep negative slope matches (y=-2x-1) (iv).
Matching the first row of graphs:
- First graph (steep positive slope, crosses y - axis at (y=-1)): (y = 2x-1) (iii)
- Second graph (horizontal line at (y=-1)): (y=-1) (ix)
- Third graph (steep negative slope, passes through origin): (y=-2x) (vi)
Matching the second row of graphs:
- First graph (steep negative slope, crosses y - axis at (y = 1)): (y=-2x + 1) (ii)
- Second graph (steep positive slope, passes through origin): (y = 2x) (v)
- Third graph (horizontal line at (y = 2)): (y = 2x) (vii)
Matching the third row of graphs:
- First graph (steep positive slope, crosses y - axis at (y = 1)): (y = 2x+1) (i)
- Second graph (horizontal line at (y = 1)): (y = 1) (viii)
- Third graph (steep negative slope, crosses y - axis at (y=-1)): (y=-2x-1) (iv)
Answer:
First row:
- First graph: (iii) (y = 2x-1)
- Second graph: (ix) (y=-1)
- Third graph: (vi) (y=-2x)
Second row:
- First graph: (ii) (y=-2x + 1)
- Second graph: (v) (y = 2x)
- Third graph: (vii) (y = 2)
Third row:
- First graph: (i) (y = 2x+1)
- Second graph: (viii) (y = 1)
- Third graph: (iv) (y=-2x-1)