Answer:
B) [tex]y=-(x-3)^{2}+2[/tex]
Step-by-step explanation:
We are given,
The graph of the parabola opens downwards.
This means that the leading co-efficient of the function will be negative.
So, options C and D are not possible.
Moreover, it is given that the graph cuts x-axis near [tex]1\frac{1}{2}[/tex] i.e. [tex]\frac{3}{2}[/tex] = 1.5 and [tex]4\frac{1}{2}[/tex] i.e. [tex]\frac{9}{2}[/tex] = 4.5.
i.e. At y=0, the value of x is near [tex]\frac{3}{2}[/tex] and [tex]\frac{9}{2}[/tex].
A. So, in [tex]y=-(x+2)^{2}+2[/tex], we put y =0.
i.e. [tex](x+2)^{2}=2[/tex]
i.e. [tex]x=-2\pm \sqrt{2}[/tex]
i.e. x = 0.68 and x = 3.41
We see that this does not crosses the x-axis at the given points.
So, option A is wrong.
B. Also, in [tex]y=-(x-3)^{2}+2[/tex], we put y =0.
i.e. [tex](x-3)^{2}=2[/tex]
i.e. [tex]x=3\pm \sqrt{2}[/tex]
i.e. x = 1.68 and x = 4.41.
Thus, this equation cuts the x-axis near the given points i.e. 1.5 and 4.5.
Hence, the equation of the given graph is [tex]y=-(x-3)^{2}+2[/tex].
