Question

Can someone help me on this please ​

Answer 1

There is one real and repeated solution to the equation f(x)=0. There are one real and equal solutions. The equation for the function is
`\(f(x) = -2(x - 2)^2 + 2\)`. The solutions are
`\(x = 3\)` and
`\(x = 1\)`.

The graph of the quadratic function is a declining curve. The equation f(x) = 0 represents the values of x for which the function equals zero.

Type of solutions: Since the graph is a quadratic curve that touches the x-axis at a single point (the vertex), the solutions are real and equal.

Number of solutions: As the line intersects the x-intercept once, there is one real and equal solution.

The equation for the Function:

Given that the vertex of the graph is at (2, 2), we can use the vertex form of a quadratic equation, which is
`\(y = a(x - h)^2 + k\)`, where
`\((h, k)\)` is the vertex of the parabola.

Plug in the values:

`\[ f(x) = a(x - 2)^2 + 2 \]`

To find the value of a, use one of the points on the graph, for example, (0, -6):

`\[ -6 = a(0 - 2)^2 + 2 \]`

Solve for a:

`\[ -6 = 4a + 2 \]`

`\[ 4a = -8 \]`

`\[ a = -2 \]`

So, the equation for the function is
`\(f(x) = -2(x - 2)^2 + 2\)`.

Solution(s) of the Equation
`\(f(x) = 0\)`:

Set f(x) equal to 0 and solve for x:

`\[ -2(x - 2)^2 + 2 = 0 \]`

`\[ -2(x - 2)^2 = -2 \]`

`\[ (x - 2)^2 = 1 \]`

`\[ x - 2 = \pm 1 \]`

`\[ x = 3 \text{ or } x = 1 \]`

So, the solutions are
`\(x = 3\)` and
`\(x = 1\)`.