Question

Use the image to answer the question.

A downward-opening parabola is graphed on a coordinate plane. The x-axis for Width ranges from negative 2 to 10 in increments of 2 and the y-axis for Area ranges from negative 2 to 18 in increments of 2.

The graph represents the area of a rectangle whose perimeter is 16. The graph is modeled by a quadratic equation of the form f(x)=ax2+bx+c
. Given that c=0
, using the points (3,15)
and (6,12)
, which system of linear equations can be used to identify the coefficients a and b?

(1 point)
Responses

225a+15b=3
and 36a+6b=12
255 a plus 15 b equals 3 and 36 a plus 6 b equals 12

9a+3b=15
and 36a+6b=12
9 a plus 3 b equals 15 and 36 a plus 6 b equals 12

9a+3b=15
and 144a+12b=6
9 a plus 3 b equals 15 and 144 a plus 12 b equals 6

225a+15b=3
and 144a+12b=6
255 a plus 15 b equals 3 and 144 a plus 12 b equals 6
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Answer 1

To find the coefficients \( a \) and \( b \) for the quadratic equation \( f(x) = ax^2 + bx + 0 \) (since \( c = 0 \)), we will use the points provided: \( (3, 15) \) and \( (6, 12) \).

1. For the point \( (3, 15) \):
\[
f(3) = a(3^2) + b(3) = 15
\]
This simplifies to:
\[
9a + 3b = 15 \tag{1}
\]

2. For the point \( (6, 12) \):
\[
f(6) = a(6^2) + b(6) = 12
\]
This simplifies to:
\[
36a + 6b = 12 \tag{2}
\]

Now we have a system of linear equations:

1. \( 9a + 3b = 15 \)
2. \( 36a + 6b = 12 \)

Now, let's check the options:

- **Option 1:** \( 225a + 15b = 3 \) and \( 36a + 6b = 12 \)
- **Option 2:** \( 9a + 3b = 15 \) and \( 36a + 6b = 12 \) (This matches our derived equations)
- **Option 3:** \( 9a + 3b = 15 \) and \( 144a + 12b = 6 \)
- **Option 4:** \( 225a + 15b = 3 \) and \( 144a + 12b = 6 \)

The correct choice is **Option 2:**
\[
9a + 3b = 15 \quad \text{and} \quad 36a + 6b = 12
\]