How can you find the x-intercepts of a parabola with a vertex at (1, 9) and a y-intercept at (0, 6)?

To find the x-intercepts of the parabola with a vertex at (1, 9) and a y-intercept at (0, 6), you’ll need to start by using the general form of a parabola’s equation. The vertex form of a parabola is given by:

y = a(x – h)² + k

Here, (h, k) is the vertex of the parabola. For this case, plugging in the vertex (1, 9), we have:

y = a(x – 1)² + 9

Next, we can use the given y-intercept at (0, 6) to find the coefficient ‘a’. By substituting x = 0 and y = 6 into the equation:

6 = a(0 – 1)² + 9

Now, simplify:

6 = a(1) + 9

6 = a + 9

Subtracting 9 from both sides gives:

a = 6 – 9

a = -3

Now we have the complete equation of the parabola:

y = -3(x – 1)² + 9

To find the x-intercepts, we need to set y to 0:

0 = -3(x – 1)² + 9

Rearranging gives us:

3(x – 1)² = 9

Dividing both sides by 3:

(x – 1)² = 3

Next, take the square root of both sides:

x – 1 = ±√3

Solving for x gives:

x = 1 + √3 or x = 1 – √3

Thus, the x-intercepts of the parabola are:

(1 + √3, 0) and (1 – √3, 0).

In numerical form, these approximate to:

(2.73, 0) and (-0.73, 0).

So, to summarize, the x-intercepts of the parabola with the given vertex and y-intercept are (1 + √3, 0) and (1 – √3, 0).

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