We have been given an equation [tex]y=-3x^2+1[/tex]. We are asked to predict the graph of our given function.
We can see that our given equation is a quadratic function.
We know that vertex for of parabola is [tex]y=a(x-h)^2+k[/tex], where a is leading coefficient and point (h,k) is vertex of parabola.
We can rewrite our given equation in vertex form as:
[tex]y=-3(x-0)^2+1[/tex]
Upon comparing our given equation with vertex form, we can see that leading coefficient is [tex]-3[/tex] and vertex is at point [tex](0,1)[/tex].
Since vertex is at [tex](0,1)[/tex], so line of symmetry will be [tex]x=0[/tex] that is equation of y-axis. So parabola will be symmetric to y-axis.
Since leading coefficient is negative, therefore, our parabola will be downward opening and option 'b' is the correct choice.