Question

a parabola has an axis of symmetry x= 3 and passes through the point (6,-4) find another point that lies on the graph of the parabola

Answer 1

Given the parabola's axis of symmetry at x = 3 and passing through (6, -4), the equation is
`\(y = -(7)/(3)(x - 3)^2 + 3\)`. This form allows finding additional points by substituting x-values.

To determine another point on the parabola with an axis of symmetry at x = 3 and passing through the point (6, -4), we use the vertex form of a parabola:
`\(y = a(x - h)^2 + k\)`, where (h, k) is the vertex. Given the axis of symmetry as x = 3, the vertex is (3, k).

Substituting the known point (6, -4) into the equation:

`\[-4 = a(6 - 3)^2 + k\]`

Simplifying, we get:

`\[-4 = 9a + k\]`

Since we already know the vertex is (3, k), we substitute 3 for k:

`\[-4 = 9a + 3\]`

Solving for a:

`\[a = -(7)/(3)\]`

Now that we have the value of a, we can substitute it back into the vertex form equation:

`\[y = -(7)/(3)(x - 3)^2 + 3\]`

This gives us the equation of the parabola. If needed, we can find additional points on the graph by substituting different x-values and calculating the corresponding y-values.