Asked by hi

Use the image to answer the question.

A coordinate plane's x-axis ranges from negative 6 to 6 and its y-axis ranges from negative 5 to 5, both by 1-unit increments. A concave down parabola is drawn passing through the 1st, 3rd, and 4th quadrants.

Write the coordinates of the vertex for the quadratic equation in the image.

(1 point)
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Answers

There are no human answers yet.
Answered by GPT-4o mini
To provide you with the coordinates of the vertex of the parabola, I would need to see the image you're referring to, as I don't have the capability to view images. However, I can guide you on how to find the vertex of a parabola.

If the parabola opens downwards, its vertex will be the highest point on the graph. You can find the vertex of a quadratic function in the standard form \( y = ax^2 + bx + c \) using the formula for \( x \):

\[ x = -\frac{b}{2a} \]

Once you have the x-coordinate, you can substitute it back into the equation to find the y-coordinate.

If you have more information about the quadratic equation or the coordinates of the points where the parabola passes through, please share it, and I can help you find the vertex. Otherwise, to get the vertex from the image, look for the highest point of the parabola on the graph.

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