Question

The parabola has a vertex at (-2, -6), two x-intercepts at (3.5, 0), and (-0.5, 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 determine the constant term \( c \) of the quadratic function in standard form \( y = ax^2 + bx + c \), we can analyze the information provided:

1. The x-intercepts are \( (3.5, 0) \) and \( (-0.5, 0) \). This implies that the quadratic can be expressed in factored form:
\[
y = a(x - 3.5)(x + 0.5)
\]

2. The vertex of the parabola is at \( (-2, -6) \). This corresponds to the point where the parabola reaches its maximum or minimum value.

3. The y-intercept is at \( (0, 6) \). The y-intercept can be found by evaluating the function at \( x = 0 \):
\[
y(0) = c
\]

### Step 1: Find \( a \) using the x-intercepts and vertex

From the expression \( y = a(x - 3.5)(x + 0.5) \), we can expand it:
\[
y = a(x^2 - 3x - 1.75)
\]

Setting up the equation around the vertex at \( x = -2 \), we can also express the vertex form:
\[
y = a(x + 2)^2 - 6
\]

### Step 2: Find the value of \( a \)

To find \( a \), we can substitute one of the x-intercepts into either the factored form or vertex form. Using \( (3.5, 0) \):
\[
0 = a(3.5 - 3.5)(3.5 + 0.5) = 0
\]
This does not help us with \( a \) because \( y \) equals zero there.

Instead, consider using \( (0, 6) \) in the vertex form:
\[
6 = a(0 + 2)^2 - 6
\]
\[
6 = 4a - 6
\]
Solving for \( a \):
\[
4a = 12
\]
\[
a = 3
\]

### Step 3: Write out the equation in standard form

Now substitute \( a \) back into the expanded form:
\[
y = 3(x^2 - 3x - 1.75)
\]
\[
= 3x^2 - 9x - 5.25
\]

### Step 4: Determine \( c \)

In standard form \( y = ax^2 + bx + c \):
- \( a = 3 \)
- \( b = -9 \)
- \( c = -5.25 \)

Thus, the constant term \( c \) (the value where the parabola intersects the y-axis) is:
\[
\boxed{-5.25}
\]