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secondary 3 | E Maths

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I need help with (a)

Date Posted: 1 month ago
Views: 23

### See 1 Answer

y = 4x² + 12x + 9
On the x-axis, y = 0
Sub y = 0 into the equation to find the intersection points.
4x² + 12x + 9 = 0
(2x)² + 2(2x)(3) + 3² = 0
(2x + 3)² = 0
2x + 3 = 0
2x = -3
x = -3/2
So there is only one point of intersection between the axis and the curve.
For a parabola, a straight line will always :
① intersect it at two points
② no intersection
③ one point of intersection (turning point touches the line)
In this case it is ③
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Date Posted: 1 month ago
J
1 month ago
So the coordinates of A are (-3/2, 0)

Or (-1½,0) or (-1.5,0) if you're not allowed to leave it in improper fractions.

Alternatively, complete the square.

y = 4x² + 12x + 9
y = (2x)² + 2(2x)(3x) + 3²
y = (2x + 3)²
(Recall (a + b)² = a² + 2ab + b²)

y = (2(x + 3/2))²
y = 4(x + 3/2)²
y = 4(x - (-3/2))²

This is now in the vertex form y = a(x - h)² + k, where (h,k) are the coordinates of the parabola's turning point.

So h = -3/2, k = 0

The turning point has y-coordinate 0, so this means the parabola only touches y = 0 (the x-axis) at 1 point

A(-3/2, 0)
Sheeeeesh
1 month ago
Hi why is the x-coordinate of A positive 1.5 when x is negative 1.5?
J
1 month ago
Typo. -3/2 is correct. Edited