To convert the quadratic function from vertex form to standard form, you first need to understand what both forms look like. The vertex form of a quadratic equation is typically given as:
f(x) = a(x – h)^2 + k
Where:
- a is a coefficient that affects the width and direction of the parabola.
- (h, k) is the vertex of the parabola.
Your question appears to contain a small error, as the function you provided seems to be in standard form already: f(x) = x^2 + 32x + 1. However, I can guide you through converting it back to vertex form and then showing you how it can be represented in standard form:
1. **Identify the coefficients**: From the function, f(x) = x^2 + 32x + 1, we can identify:
- a = 1
- b = 32
- c = 1
2. **Finding the vertex**: The formula for the x-coordinate of the vertex (h) is given by:
h = -b/(2a) = -32/(2 * 1) = -16
The y-coordinate (k) can be found by substituting h back into the function:
k = f(-16) = (-16)^2 + 32*(-16) + 1 = 256 – 512 + 1 = -255
So the vertex is at the point (-16, -255).
3. **Writing the vertex form**: Now, we can write the vertex form as:
f(x) = 1(x + 16)^2 – 255
4. **Expanding to standard form**: To convert from vertex form back to standard form, we can expand:
First, expand the squared term:
(x + 16)^2 = x^2 + 32x + 256
Now substitute back into the equation:
f(x) = 1(x^2 + 32x + 256) – 255
5. **Combine like terms**:
f(x) = x^2 + 32x + 256 – 255
f(x) = x^2 + 32x + 1
Thus, both forms are equivalent, confirming that your function is already represented in standard form as f(x) = x^2 + 32x + 1.
In summary, regardless of the starting point, both the vertex form and standard form are interconnected, and with some calculations, you can navigate between them smoothly.