Answer:
The function which represents the graph is:
[tex]y=-\left(x\:-\:2\right)^2+\:3[/tex]
Hence, option B is correct.
The graph is also attached.
Step-by-step explanation:
From the graph, it is clear that:
The vertex = (2, 3)
The y-intercept = (0, -1)
Let us check the equation
[tex]y=-\left(x\:-\:2\right)^2+\:3[/tex]
Determining the y-intercept:
The value of the y-intercept can be determined by setting x = 0, and determining the corresponding value of y.
substituting x = 0 in teh equation
[tex]y=-\left(x\:-\:2\right)^2+\:3[/tex]
[tex]y=-\left(0\:-\:2\right)^2+\:3[/tex]
[tex]y=-2^2+3[/tex]
[tex]y=-4+3[/tex]
[tex]y=-1[/tex]
Thus,
The point representing the y-intercept is: (0, -1)
Determining the vertex:
The vertex of an up-down facing parabola of the form ax² + bx + c is:
[tex]x_v=-\frac{b}{2a}[/tex]
As the equation is
[tex]y=-\left(x\:-\:2\right)^2+\:3[/tex]
Rewriting the equation [tex]y=-\left(x\:-\:2\right)^2+\:3[/tex] in the form ax² + bx + c:
[tex]y=-x^2+4x-1[/tex]
The Parabola params are:
[tex]a=-1,\:b=4,\:c=-1[/tex]
Thus,
[tex]x_v=-\frac{b}{2a}[/tex]
[tex]x_v=-\frac{4}{2\left(-1\right)}[/tex]
[tex]x_v=2[/tex]
Now plug in [tex]x_v=2[/tex] in the equation to find [tex]y_v[/tex]
[tex]y_v=-x^2+4x-1[/tex]
[tex]\:y_v=-\left(2\right)^2+4\left(2\right)-1[/tex]
[tex]y_v=-4+8-1[/tex]
[tex]y_v=3[/tex]
Therefore, the vertex of the equation is: (2, 3)
Hence, the function which represents the graph is:
[tex]y=-\left(x\:-\:2\right)^2+\:3[/tex]
Hence, option B is correct.
The graph is also attached.
