Find the x-intercepts of the parabola with vertex (1, -9) and y intercept at (0, -6).

Solution:

Given, vertex = (1, -9)

y-intercept at (0, -6)

We have to find the x-intercepts of the parabola.

The general equation of parabola is given by

\(y=a(x-h)^{2}+k\)

Where, (h, k) is the vertex

Now, (h, k) = (1, -9)

So, \(y=a(x-1)^{2}+(-9)\)

\(y=a(x-1)^{2}-9\)

Substituting the value of y in the above equation,

\(-6=a(0-1)^{2}-9\)

\(-6=a(-1)^{2}-9\)

\(-6=a-9\)

a = -6 + 9

a = 3

Now, to find x-intercept put y = 0,

\(0=3(x-1)^{2}-9\)

\(3(x-1)^{2}=9\)

\((x-1)^{2}=\frac{9}{3}\)

\((x-1)^{2}=3\)

Taking square root,

(x - 1) = ±√3

x = 1 ± √3

Therefore, the x-intercepts are x = 1 - √3 and x = 1 + √3.

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Find the x-intercepts of the parabola with vertex (1, -9) and y intercept at (0, -6).

Summary:

The x-intercepts of the parabola with vertex (1, -9) and y-intercept at (0, -6) are x = 1 - √3 and x = 1 + √3.