Question

The parabola has a vertex at (-2, -6), a x-intercept at (3.5, 0), and a y-intercept at (0, 6).

Determine the value of the constant term of the quadratic function in standard form, given its graph.

The value of c in the standard form of the quadratic function y = ax^2 + box + c is _

Answer 1

To find the value of the constant term \( c \) in the standard form of the quadratic function \( y = ax^2 + bx + c \), we can utilize the information we have about the parabola: the vertex, x-intercept, and y-intercept.

1. **Vertex:** The vertex of the parabola is given as \( (-2, -6) \). In the vertex form of a parabola, the equation can be expressed as:

\[
y = a(x + 2)^2 - 6
\]

2. **X-intercept:** The x-intercept is at \( (3.5, 0) \). We can substitute this point into our vertex equation to solve for \( a \):

\[
0 = a(3.5 + 2)^2 - 6
\]

Simplifying, we find:

\[
0 = a(5.5)^2 - 6
\]

\[
0 = 30.25a - 6
\]

\[
30.25a = 6 \quad \Rightarrow \quad a = \frac{6}{30.25} = \frac{6}{\frac{121}{4}} = \frac{6 \cdot 4}{121} = \frac{24}{121}
\]

Now we have \( a = \frac{24}{121} \).

3. **Convert to Standard Form:** Now we can substitute \( a \) back into the vertex form to convert it to standard form:

\[
y = \frac{24}{121}(x + 2)^2 - 6
\]

Expanding this:

\[
y = \frac{24}{121}(x^2 + 4x + 4) - 6
\]

\[
y = \frac{24}{121}x^2 + \frac{96}{121}x + \frac{96}{121} - 6
\]

To combine \( \frac{96}{121} - 6 \) into a single fraction, note that \( 6 = \frac{726}{121} \):

\[
\frac{96}{121} - \frac{726}{121} = \frac{96 - 726}{121} = \frac{-630}{121}
\]

Thus, the standard form of the equation becomes:

\[
y = \frac{24}{121}x^2 + \frac{96}{121}x - \frac{630}{121}
\]

4. **Constant Term:** The constant term in the standard form \( y = ax^2 + bx + c \) is represented as \( c \).

Thus, the value of the constant term \( c \) is:

\[
c = -\frac{630}{121}
\]