Question

Given the equation y=x2−16x+89 with zeros of x=8−5i and x=8+5i, which of the following identifies the general shape of its associated graph?(1 point) Responses The graph touches the x-axis exactly twice. The graph touches the x -axis exactly twice. The graph opens downward. The graph opens downward. The vertex is to the left of the y-axis. The vertex is to the left of the y -

Answer 1

The equation you provided is a quadratic function of the form \(y = ax^2 + bx + c\). In your case, the equation is:

\[
y = x^2 - 16x + 89
\]

Given that the zeros of the function are complex (specifically \(x = 8 - 5i\) and \(x = 8 + 5i\)), this indicates that the graph of the function does not intersect the x-axis. Instead, since the zeros are complex and have no real parts, the graph opens upward because the coefficient of \(x^2\) (which is 1) is positive.

Now let's consider the responses provided:

1. The graph touches the x-axis exactly twice. (Incorrect)
2. The graph opens downward. (Incorrect)
3. The vertex is to the left of the y-axis. (Potentially correct, but we will address in a bit)

Since the graph opens upward and does not touch the x-axis, the statements about touching the x-axis are incorrect.

To determine the position of the vertex, we can find the x-coordinate of the vertex using the formula \(x = -\frac{b}{2a}\):

Here, \(a = 1\) and \(b = -16\):

\[
x = -\frac{-16}{2 \cdot 1} = \frac{16}{2} = 8
\]

The vertex is at \(x = 8\), which is on the y-axis, not to the left.

### Conclusion:
- The graph **does not touch the x-axis at all**.
- The graph **opens upward**.
- The vertex **is on the y-axis**.

Thus, it seems the most accurate response among the choices you gave would be that **the graph opens upward**, though that wasn't one of the specified choices. From what you provided, no option accurately describes the graph.