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)
Accepted Solution
A:
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.