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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

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

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)

Hi why is the x-coordinate of A positive 1.5 when x is negative 1.5?

Typo. -3/2 is correct. Edited