Question

Help me solve 1c. and 1d. please :) ​

Answer 1

The quadratic equation for the given parabola with a vertex at (2, 5) is
`\(y = -(x - 2)^2 + 5\)`. Solving for a using the point (0, 1) yields a = -1.

To write the equation for the quadratic and solve for \(a\), we would use the vertex form of a quadratic equation. The vertex form is given by:

`\[ y = a(x - h)^2 + k \]`

where (h, k) is the vertex of the parabola. In this case, you mentioned that the vertex is at (2, 5), so the equation for the quadratic would be:

`\[ y = a(x - 2)^2 + 5 \]`

To solve for a, you can substitute the coordinates of one of the points on the parabola (e.g., (0, 1) or (4, 1) into the equation and then solve for a. This process will yield the specific equation for the given quadratic.

Now, substitute the coordinates of a point on the parabola to solve for (a). Let's use the point (0, 1):

`\[ 1 = a(0 - 2)^2 + 5 \]`

Simplify the equation:

`\[ 1 = 4a + 5 \]`

Subtract 5 from both sides:

`\[ -4 = 4a \]`

Divide by 4:

`\[ a = -1 \]`

Now that we have found the value of (a), we can write the specific equation for the given quadratic:

`\[ y = -1(x - 2)^2 + 5 \]`

So, the equation for the quadratic is
`\(y = -(x - 2)^2 + 5\)`, and the value of
`\(a\)` is -1.