Question 5

Parabolic Curve

Visualization of the parabolic curve on the cartesian plane.

The correct statement regarding the curve is:

Let's analyze each statement one by one based on the given graph.

1. Checking the Y-Axis Intercept (Statement 1)

From the image, the intersection point with the $y$ -axis appears to be at $(0, -1)$ and another point at the top.

Let's find the equation first. The vertex of the parabola is at $(4, 1)$ .

The general form of a horizontal parabola equation with vertex $(x_p, y_p)$ is $x = a(y - y_p)^2 + x_p$ .

Substitute $x_p = 4$ and $y_p = 1$ :

x = a(y - 1)^2 + 4

The curve passes through the point $(0, -1)$ . We substitute this point to find the value of $a$ .

0 = a(-1 - 1)^2 + 4

0 = a(-2)^2 + 4

0 = 4a + 4

4a = -4

a = -1

So, the equation of the curve is $x = -(y - 1)^2 + 4$ .

Now we find the intersection points with the $y$ -axis (when $x = 0$ ):

0 = -(y - 1)^2 + 4

(y - 1)^2 = 4

y - 1 = \pm 2

Solution 1: $y - 1 = 2 \Rightarrow y = 3$ . The intersection point is $(0, 3)$ .

Solution 2: $y - 1 = -2 \Rightarrow y = -1$ . The intersection point is $(0, -1)$ .

Statement (1) states $(0, 2)$ , which is FALSE.

2. Checking the Curve Equation (Statement 2)

From the calculation above, we have obtained the equation of the curve is:

x = -(y - 1)^2 + 4

In function notation $f(y)$ , this is written as $f(y) = -(y - 1)^2 + 4$ .

Statement (2) is TRUE.

3. Checking the X-Axis Intercept (Statement 3)

The intersection point with the $x$ -axis occurs when $y = 0$ .

x = -(0 - 1)^2 + 4

x = -(-1)^2 + 4

x = -1 + 4

x = 3

So, the intersection point with the $x$ -axis is $(3, 0)$ .

Statement (3) states $(1, 0)$ , which is FALSE.

4. Checking the Vertex (Statement 4)

From the graph, it is clearly visible that the vertex is $(4, 1)$ .

The abscissa ( $x$ value) of the vertex is $4$ .

Statement (4) is TRUE.

Conclusion:

The correct statements are (2) and (4).

Thus, the answer is C.

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