Set y=f(x)=2x^2y=f(x)=2x2
Compare to the standardised form: y=ax^2+bx+cy=ax2+bx+c
This equation color(white)("dddddfdd") ->color(white)("ddddd")y=2x^2+0x+0 dddddfdd→dddddy=2x2+0x+0
Key points:
a>0 -> ("positive")a>0→(positive) so the graph is of form uu∪
c->c→ y-intercept =0=0
x_("vertex") -> (-1/2)xx b/acolor(white)("ddd") =color(white)("ddd") (-1/2)xx0/2color(white)("d")=color(white)("d")0xvertex→(−12)×baddd=ddd(−12)×02d=d0
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As x_("vertex")=0xvertex=0 the axis of symmetry is the y-axis
As a>0a>0 then the general shape is uu∪ with the y-axis in the middle.
Couple this with c=0c=0 and it means that the vertex is at (x,y)=(0.0)(x,y)=(0.0)