Respuesta :

Answer:

Step-by-step explanation:

I used logic and took the easy way around this as opposed to the long, drawn-out algebraic way.  I noticed right off that at x = -3 and x = -1 the y values were the same.  In the middle of those two x-values is -2, which is the vertex of the parabola with coordinates (-2, 4).  That's the h and k in the formula I'm going to use.  Then I picked a point from the table to use as my x and y in the formula I'm going to use.  I chose (0, 3) because it's easy.  The formula for a quadratic is

[tex]y=a(x-h)^2+k[/tex]

and I have everything I need to solve for a.  Filling in my h, k, x, and y:

[tex]3=a(0-(-2))^2+4[/tex]  and

[tex]3=a(2)^2+4[/tex]  and

-1 = 4a so

[tex]a=-\frac{1}{4}[/tex]

In work/vertex form the equation for the quadratic is

[tex]y=-\frac{1}{4}(x+2)^2+4[/tex]

In standard form it's:

[tex]y=-\frac{1}{4}x^2-x+3[/tex]

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