Answer:
[tex]y\leq 2x^2-8x+3[/tex]
Step-by-step explanation:
we know that
The quadratic function of the figure has a vertex at (2,-5) and the y-intercept is the point (0,3)
The equation of a vertical parabola in vertex form is equal to
[tex]y=a(x-h)^2+k[/tex]
where
(h,k) is the vertex
a is a coefficient
we have
[tex](h,k)=(2,-5)[/tex]
substitute
[tex]y=a(x-2)^2-5[/tex]
with the y-intercept find out the value of a
For x=0,y=3
[tex]3=a(0-2)^2-5[/tex]
[tex]3=4a-5[/tex]
[tex]4a=8[/tex]
[tex]a=2[/tex]
The equation is
[tex]y=2(x-2)^2-5[/tex]
[tex]y=2(x^2-4x+4)-5[/tex]
[tex]y=2x^2-8x+8-5[/tex]
[tex]y=2x^2-8x+3[/tex]
Looking at the graph, the solution is the shaded area below the quadratic equation
so
the value of y must be less than or equal to the quadratic function
[tex]y\leq 2x^2-8x+3[/tex]
