Questions: Consider the following function. a(x)=(x-3)^2-1 Find two points on the graph of the parabola other than the vertex and x-intercepts.

Solution Steps

The given function is

\[ a(x) = (x - 3)^2 - 1 \]

This represents a parabola that opens upwards with its vertex at the point \((3, -1)\).

To find the x-intercepts, we set \(a(x) = 0\):

\[ (x - 3)^2 - 1 = 0 \]

Solving this gives:

\[ (x - 3)^2 = 1 \]

Taking the square root of both sides:

\[ x - 3 = \pm 1 \]

Thus, we find:

\[ x = 4 \quad \text{and} \quad x = 2 \]

The x-intercepts are at the points \((2, 0)\) and \((4, 0)\).

We select x-values that are not the vertex or the x-intercepts. We choose \(x = 1\) and \(x = 5\) and calculate the corresponding y-values:

1. For \(x = 1\):

\[ a(1) = (1 - 3)^2 - 1 = 4 - 1 = 3 \]

Thus, the point is \((1, 3)\).

2. For \(x = 5\):

\[ a(5) = (5 - 3)^2 - 1 = 4 - 1 = 3 \]

Thus, the point is \((5, 3)\).

Final Answer