Question

I'm studying quadratic equations in my math class and I need help graphing the function f(x)=2x^2+4x-8. Can you find the vertex of this parabola and plot another point on the graph?

Answer 1

Certainly! To analyze the quadratic function \( f(x) = 2x^2 + 4x - 8 \), we will first find the vertex and then another point to help with graphing.

### Step 1: Find the Vertex

The vertex of a parabola described by the equation \( f(x) = ax^2 + bx + c \) can be found using the formula for the x-coordinate of the vertex:

\[
x = -\frac{b}{2a}
\]

In this case:
- \( a = 2 \)
- \( b = 4 \)

Plugging in the values:

\[
x = -\frac{4}{2 \cdot 2} = -\frac{4}{4} = -1
\]

Now, we will find the y-coordinate of the vertex by substituting \( x = -1 \) back into the function:

\[
f(-1) = 2(-1)^2 + 4(-1) - 8
\]
\[
= 2(1) - 4 - 8
\]
\[
= 2 - 4 - 8
\]
\[
= -10
\]

So, the vertex of the parabola is at the point \( (-1, -10) \).

### Step 2: Find Another Point

Let’s choose another simple x-value, like \( x = 0 \), and find the corresponding y-value:

\[
f(0) = 2(0)^2 + 4(0) - 8
\]
\[
= 0 + 0 - 8
\]
\[
= -8
\]

So, another point on the graph is \( (0, -8) \).

### Summary of Points

- Vertex: \( (-1, -10) \)
- Another point: \( (0, -8) \)

### Graphing

To graph the function, we can plot the points \( (-1, -10) \) and \( (0, -8) \). Since the parabola opens upwards (because \( a = 2 > 0 \)), it will form a U-shape.

1. Plot the vertex \( (-1, -10) \).
2. Plot the point \( (0, -8) \).
3. You can continue to plot additional points if required, for example, using \( x = 1 \):
\[
f(1) = 2(1)^2 + 4(1) - 8 = 2 + 4 - 8 = -2
\]
This yields another point \( (1, -2) \).

Now you can sketch the parabola using these points, showing how it curves upwards around the vertex.