The graph of $y = 2x^2$ is passing through the point : (a) $(2, 0)$ (b) $(0, 0)$ (c) $(0, 2)$ (d) $(2, 1)$

The graph of y = 2x² is passing through the point : (a) (2, 0) (b) (0, 0) (c) (0, 2) (d) (2, 1)

Question 33 of 49

y = 2x^2

(a)

(2, 0)

(b)

(0, 0)

(c)

(0, 2)

(d)

(2, 1)

Puducherry TnboardTamil Nadu HSC First Year (DGE) Commerce Board 2023MCQ· 1mImportance★★★★★

67% · 33/49 Questions

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The point $(0,0)$ satisfies $y = 2x^2$ (it is the vertex), so the answer is option (b).

A point lies on the curve $y = 2x^2$ only if its coordinates satisfy the equation. Check the vertex point $(0, 0)$ :

$y = 2x^2 = 2(0)^2 = 0,$

which matches $y = 0$ . So $(0,0)$ lies on the graph.

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