Question

Use the image to answer the question.

A coordinate plane shows a parabola that opens downward with a plotted point. Both axes range from negative 8 to 8 in increments of 2. the plotted point is (-3,-1)

What is the vertex form of the graphed equation?

(1 point)
y=−2(x−_)^2 +_
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Answer 1

To determine the vertex form of the equation of the parabola, we need to identify the vertex coordinates.

Given the plotted point (-3, -1), we can assume that this is the vertex of the parabola, based on the information provided. The vertex form of a parabola is represented as:

\[ y = a(x - h)^2 + k \]

where \((h, k)\) is the vertex of the parabola. In your case, since the vertex is at (-3, -1):

- \(h = -3\)
- \(k = -1\)

Now, the only thing we need to find is the value of \(a\). We see that the parabola opens downward, which indicates that \(a\) will be negative.

Let's assume \(a = -2\) as indicated by your question, making our equation look like this:

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

Thus, the completed form is:

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

So, filling the blanks in the form \(y = −2(x−_)^2 + _\):

- \( _ = -3 \)
- \( _ = -1 \)

In summary, the vertex form of the graphed equation is:

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

And the values for the blanks would be:

**(-3, -1)** in the vertex form answer.