Question

Identify the vertex, axis of symmetry, y-intercept, and zeros of the quadratic equation. Drag and drop the answers into the correct box.2x2−12x+16

Answer 1

Let's analyze the quadratic equation given: \( f(x) = 2x^2 - 12x + 16 \).

1. **Vertex**: The vertex of a parabola defined by the equation in the form \( ax^2 + bx + c \) can be found using the vertex formula \( x = -\frac{b}{2a} \).

Here, \( a = 2 \) and \( b = -12 \):
\[
x = -\frac{-12}{2 \cdot 2} = \frac{12}{4} = 3
\]

To find the y-coordinate of the vertex, plug \( x = 3 \) back into the equation:
\[
f(3) = 2(3)^2 - 12(3) + 16 = 2(9) - 36 + 16 = 18 - 36 + 16 = -2
\]
So, the vertex is \( (3, -2) \).

2. **Axis of Symmetry**: The axis of symmetry is the vertical line that passes through the vertex. Thus, the axis of symmetry is:
\[
x = 3
\]

3. **Y-Intercept**: The y-intercept can be found by evaluating \( f(0) \):
\[
f(0) = 2(0)^2 - 12(0) + 16 = 16
\]
So, the y-intercept is \( (0, 16) \).

4. **Zeros**: To find the zeros of the equation, set \( f(x) = 0 \) and solve for \( x \):
\[
2x^2 - 12x + 16 = 0
\]
We can simplify this by dividing everything by 2:
\[
x^2 - 6x + 8 = 0
\]
This factors as:
\[
(x - 2)(x - 4) = 0
\]
Thus, the zeros are:
\[
x = 2 \quad \text{and} \quad x = 4
\]

Now putting everything together, we have:

- **Vertex**: \( (3, -2) \)
- **Axis of Symmetry**: \( x = 3 \)
- **Y-Intercept**: \( (0, 16) \)
- **Zeros**: \( 2 \) and \( 4 \)

You would drag and drop these items into their respective boxes as outlined.