Use the image to answer the question.

Question

A parabola is graphed on a coordinate plane. The x-axis ranges from negative 6 to 6 in increments of 1. The y-axis ranges from negative 6 to 6 in increments of 1.

Write the equation of the quadratic function in the graph in standard form.

(2 points)
The standard form of the quadratic is y=_x2+_x+_

On the graph, it curves up from -4 and curves down at 2

Answer 1

To determine the standard form of the quadratic function for the parabola described, we can use the information given.

You've mentioned that the parabola opens upward from the vertex at \( y = -4 \) and has a point where it "curves down" at \( x = 2 \). This suggests that the parabola has a minimum point (the vertex) and the vertex form can be utilized.

The vertex form of a quadratic function is given by:

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

where \( (h, k) \) is the vertex of the parabola.

From the information provided:

• The vertex is at \( (h, k) = (2, -4) \).

So we have:

\[ y = a(x - 2)^2 - 4 \]

To find the value of \( a \), we need another point on the parabola. If the description suggests that the parabola "curves down" at 2, we can evaluate the standard form using \( y \) coordinates higher than -4 (the value of the vertex).

Assuming the parabola crosses y-axis at some point, for instance, the point \( (0, 0) \):

Substituting \( x = 0 \) and \( y = 0 \):

\[ 0 = a(0 - 2)^2 - 4 \]

\[ 0 = a(4) - 4 \]

\[ 4 = 4a \]

\[ a = 1 \]

Thus, we can plug \( a = 1 \) back into the vertex form:

\[ y = (x - 2)^2 - 4 \]

Expanding this to standard form:

\[ y = (x^2 - 4x + 4) - 4 \]

\[ y = x^2 - 4x + 0 \]

So, the standard form of the quadratic is:

\[ y = 1x^2 - 4x + 0 \]

In summary:

The standard form of the quadratic is \( y = 1x^2 - 4x + 0 \).