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secondary 3 | A Maths
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Do help me out! I don’t know how to continue and if my working is even correct
Date Posted: 5 years ago
Views: 256
The completing the square part looks ok. To verify, just re-expand the factorised term.
The value of y depends on the value of x. Of course, we know how low (x - 2)^2 can go.
So to find the minimum point, we set x - 2 = 0. We get x = 2, and the corresponding minimum value of y is -1. So the minimum point is (2, -1).
The y-intercept is (0, 3).
Since the coefficient of x2 is positive, we draw a 'smiley face' graph.
The value of y depends on the value of x. Of course, we know how low (x - 2)^2 can go.
So to find the minimum point, we set x - 2 = 0. We get x = 2, and the corresponding minimum value of y is -1. So the minimum point is (2, -1).
The y-intercept is (0, 3).
Since the coefficient of x2 is positive, we draw a 'smiley face' graph.
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Something like this would be good.
We just need to find the coordinates of the turning point, the coordinates of the y-intercept and the signage of the coefficient of x2 to draw the graph. The graph will always be symmetrical about the vertical line passing through the turning point.
To be safe, we can find the x-intercepts also where possible (though not necessary all the time).
We just need to find the coordinates of the turning point, the coordinates of the y-intercept and the signage of the coefficient of x2 to draw the graph. The graph will always be symmetrical about the vertical line passing through the turning point.
To be safe, we can find the x-intercepts also where possible (though not necessary all the time).
Date Posted:
5 years ago
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