Question

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)

Answer 1

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,
$x=\frac{-(-1)\pm\sqrt{(-1)^2-4(0.25)(-3)}}{2(0.25)} \\ \\=\frac{1\pm \sqrt{1--3}}{0.5}=\frac{-1\pm \sqrt{4}}{0.5} \\ \\=\frac{1\pm2}{0.5}=\frac{1+2}{0.5}\text{ or }\frac{1-2}{0.5} \\ \\=\frac{3}{0.5}\text{ or }\frac{-1}{0.5}=6\text{ or }-2$

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

Answer 2

The value of $\text{g}$ is $\boxed6$ and the cut point is $\boxed{\left( {6,0} \right)}$.

Further Explanation:

A polynomial with degree $2$ is called as quadratic function.

The general quadratic function with vertex $\left( {h,k} \right)$ can be expressed as,

$\boxed{y - k = a{{\left( {x - h} \right)}^2}}$

Given:

The vertex of the quadratic function is $\left( {2, - 4} \right)$.

The cut points on the $x$-axis are $\left( { - 2,0} \right)$ and $\left( {\text{g},0} \right)$.

Calculation:

Substitute $2$ for $h$ and $-4$ for $k$ in general quadratic equation.

$\begin{aligned}y - \left( { - 4} \right) &= a{\left( {x - 2} \right)^2} \\y + 4 &= a{\left( {x - 2} \right)^2} \\y &= a{\left( {x - 2} \right)^2} - 4 \\\end{aligned}$

The passing through point is $\left( { - 2,0} \right)$.

Substitute $-2$ for $x$ and $0$ for $y$ in equation $y = a{\left( {x - 2} \right)^2} -4$.

$\begin{aligned}0&= a{\left( { - 2 - 2} \right)^2} - 4 \\4 &= 16a \\\frac{4}{{16}} &= a \\0.25 &= a \\\end{aligned}$

Now the quadratic function can be expressed as,

$\boxed{y = 0.25{{\left( {x - 2} \right)}^2} - 4}$

The passing through point is $\left( { - 2,0} \right)$.

Substitute $\text{g}$ for $x$ and 0 for $y$ in equation $y = 0.25{{\left( {x - 2} \right)}^2}-4$.

$\begin{aligned}0 &= 0.25{\left( {\text{g} - 2} \right)^2} - 4 \\4 &= 0.25{\left( {\text{g} - 2} \right)^2} \\\frac{4}{{0.25}} &= {\left( {\text{g} - 2} \right)^2} \\16 &= {\left( {\text{g} - 2} \right)^2} \\{\left( 4 \right)^2} &= {\left( {\text{g} - 2} \right)^2} \\\end{aligned}$

The value of $\text{g-2}$ is equal to $4$.

Then the value of $\text{g}$ can be obtained as,

$\begin{aligned}\text{g}-2 &= 4 \\\text{g}&= 4 + 2\\&= 6 \\\end{aligned}$

The value of $\text{g}$ is $\boxed6$ and the cut point is $\boxed{\left( {6,0} \right)}$.

The quadratic function is $\boxed{y = 0.25{{\left( {x - 2} \right)}^2} - 4}$

Learn more:

1. Learn more about unit conversion https://brainly.com/question/4837736

2. Learn more about non-collinear https://brainly.com/question/4165000

Answer details:

Grade: Middle School

Subject: Mathematics

Chapter: Quadratic Equations

Keywords: sum, positive number, quadratic equation, minus, difference, square root, number, value, vertex, cut of points, represents, standard quadratic equation.