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secondary 3 | A Maths
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The expression hx² + 6x + k is in the form of the equation of a parabola y = ax² + bx + c (or quadratic equation if you like).
Since the maximum value is negative (< 0) , this tells you two things :
① The curve/parabola is ∩ shaped and downward sloping. (Its maximum point/value is at the apex/vertex of the curve)
② The entire curve/parabola is below the x-axis (where y = 0). There are no intersection points with the x-axis.
Based on ① , we can say that the coefficient of x² must be negative. Looking at the original expression, we can deduce that h is negative (h < 0)
Based on ②, we can deduce that there are no real roots when hx² + 6x + k = 0 (i.e The curve y = hx² + 6x + k has no points of intersection with the line y = 0)
So the discriminant (b² - 4ac) < 0
6² - 4hk < 0
36 < 4hk
9 < hk
Since we know h < 0, then 9/h > k (we need to flip the sign when dividing by a negative value)
Since the maximum value is negative (< 0) , this tells you two things :
① The curve/parabola is ∩ shaped and downward sloping. (Its maximum point/value is at the apex/vertex of the curve)
② The entire curve/parabola is below the x-axis (where y = 0). There are no intersection points with the x-axis.
Based on ① , we can say that the coefficient of x² must be negative. Looking at the original expression, we can deduce that h is negative (h < 0)
Based on ②, we can deduce that there are no real roots when hx² + 6x + k = 0 (i.e The curve y = hx² + 6x + k has no points of intersection with the line y = 0)
So the discriminant (b² - 4ac) < 0
6² - 4hk < 0
36 < 4hk
9 < hk
Since we know h < 0, then 9/h > k (we need to flip the sign when dividing by a negative value)
Alternatively,
We can complete the square to get the vertex of the parabola.
Let y = hx² + 6x + k
= h(x² + 6x/h) + k
= h(x² + 2(1)(3x/h) + (3/h)² - (3/h)²) + k
= h((x + 3/h)² - 9/h²) + k
= h(x + 3/h)² - 9h/h² + k
= h(x + 3/h)² + k - 9/h
= h(x - (-3/h))² + (k - 9/h)
The coordinates of the vertex of the parabola are (-3/h, k - 9/h)
Since the maximum point/value of hx² + 6x + k is negative,
① h < 0 in order to have a ∩-shaped downward sloping curve.
② The y-coordinate of the vertex < 0
So k - 9/h < 0
k < 9/h
We can complete the square to get the vertex of the parabola.
Let y = hx² + 6x + k
= h(x² + 6x/h) + k
= h(x² + 2(1)(3x/h) + (3/h)² - (3/h)²) + k
= h((x + 3/h)² - 9/h²) + k
= h(x + 3/h)² - 9h/h² + k
= h(x + 3/h)² + k - 9/h
= h(x - (-3/h))² + (k - 9/h)
The coordinates of the vertex of the parabola are (-3/h, k - 9/h)
Since the maximum point/value of hx² + 6x + k is negative,
① h < 0 in order to have a ∩-shaped downward sloping curve.
② The y-coordinate of the vertex < 0
So k - 9/h < 0
k < 9/h
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