Representa una funcion cuadratica con las siguientes caracteristicas a: vertice en (2,-4)y b: puntos de corte con el eje x(-2,0)y (g,0)

The quadratic function is given by the equation
y=0.25(x-2)²-4, and the value of g is 6.

We first write this in vertex form, using the information we have:
y=a(x-h)²+k, where (h, k) is the vertex
y=a(x-2)²-4

We now substitute one of the points the function goes through into our x and y variables to solve for a:
0=a(-2-2)²-4
0=a(-4)²-4
0=16a-4

Add 4 to both sides:
0+4 = 16a-4+4
4=16a

Divide both sides by 16:
4/16 = 16a/16
0.25 = a

This gives us the function
y=0.25(x-2)²-4

Now we write this in standard form:
y=0.25(x-2)(x-2)-4
y=0.25(x*x-2*x-2*x-2(-2))-4
y=0.25(x²-2x-2x+4)-4
y=0.25(x²-4x+4)-4
y=0.25x²-1x+1-4
y=0.25x²-1x-3

Using the quadratic formula,
[tex]x=\frac{-(-1)\pm\sqrt{(-1)^2-4(0.25)(-3)}}{2(0.25)} \\ \\=\frac{1\pm \sqrt{1--3}}{0.5}=\frac{-1\pm \sqrt{4}}{0.5} \\ \\=\frac{1\pm2}{0.5}=\frac{1+2}{0.5}\text{ or }\frac{1-2}{0.5} \\ \\=\frac{3}{0.5}\text{ or }\frac{-1}{0.5}=6\text{ or }-2[/tex]

We already had the root (-2, 0); this gives us the value of g, 6.