Question

Khalid drew the graph of the quadratic function f(x)=2x2+6x+4.5
shown in the graph. What is the range of this function?

(1 point)
Responses

y≥0
y is greater than or equal to 0

y≤0
y is less than or equal to 0

y>0
y is greater than 0

−1.5≤y≤0

Answer 1

To find the range of the quadratic function \( f(x) = 2x^2 + 6x + 4.5 \), we can start by finding the vertex, since the vertex gives us the minimum (or maximum) point of the parabola.

The quadratic function can be expressed in the form \( ax^2 + bx + c \), where:
- \( a = 2 \)
- \( b = 6 \)
- \( c = 4.5 \)

Since \( a \) is positive, the parabola opens upward, and thus the vertex will represent the minimum value of the function.

The x-coordinate of the vertex can be found using the formula:
\[
x = -\frac{b}{2a}
\]
Substituting in the values:
\[
x = -\frac{6}{2 \times 2} = -\frac{6}{4} = -1.5
\]

Now we calculate the y-coordinate of the vertex by substituting \( x = -1.5 \) back into the function:
\[
f(-1.5) = 2(-1.5)^2 + 6(-1.5) + 4.5
\]
Calculating it step-by-step:
\[
= 2(2.25) + 6(-1.5) + 4.5
\]
\[
= 4.5 - 9 + 4.5
\]
\[
= 4.5 + 4.5 - 9 = 9 - 9 = 0
\]

So the vertex of the parabola is at the point \((-1.5, 0)\). Since it opens upward, this means the minimum value of \( f(x) \) is 0.

Thus, the range of the function \( f(x) = 2x^2 + 6x + 4.5 \) is:
\[
y \geq 0
\]

Therefore, the answer is:
**y≥0** (y is greater than or equal to 0)