Question

How do you graph a quadratic equation?

Answer 1

Graphing a quadratic equation involves several steps. A standard form of a quadratic equation is given by:

\[ y = ax^2 + bx + c \]

where \(a\), \(b\), and \(c\) are constants, and \(a \neq 0\). Here's how to graph a quadratic equation step by step:

### Steps to Graph a Quadratic Equation

1. **Identify the coefficients**: Determine the values of \(a\), \(b\), and \(c\) from the equation.

2. **Find the vertex**: The vertex of a parabola represented by the equation can be found using the formula:
\[
x = -\frac{b}{2a}
\]
Substitute this \(x\) value back into the equation to find the \(y\) value of the vertex:
\[
y = a\left(-\frac{b}{2a}\right)^2 + b\left(-\frac{b}{2a}\right) + c
\]
So, the vertex is located at the point \((-\frac{b}{2a}, y)\).

3. **Determine the axis of symmetry**: The line \(x = -\frac{b}{2a}\) is the axis of symmetry for the parabola.

4. **Find the y-intercept**: The y-intercept occurs where \(x = 0\). Substitute \(x = 0\) into the equation:
\[
y = c
\]
So, the y-intercept is at the point \((0, c)\).

5. **Find the x-intercepts (if any)**: Set \(y = 0\) and solve the quadratic equation:
\[
ax^2 + bx + c = 0
\]
You can use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
The discriminant (\(b^2 - 4ac\)) will determine the nature of the x-intercepts:
- If it's positive, there are two distinct real roots.
- If it's zero, there is one real root (the vertex touches the x-axis).
- If it's negative, there are no real roots (the parabola does not touch the x-axis).

6. **Plot the points**: Begin by plotting the vertex, y-intercept, and x-intercepts on the graph. If needed, calculate additional points on either side of the vertex to get a clearer shape of the parabola.

7. **Draw the parabola**: Sketch a smooth curve through the points you plotted. Parabolas open upwards if \(a > 0\) and downwards if \(a < 0\).

### Example

Let's graph the quadratic equation:

\[ y = 2x^2 - 4x + 1 \]

1. **Identify coefficients**: \(a = 2\), \(b = -4\), \(c = 1\).

2. **Find the vertex**:
\[
x = -\frac{-4}{2 \times 2} = \frac{4}{4} = 1
\]
Substitute \(x = 1\) into the equation:
\[
y = 2(1)^2 - 4(1) + 1 = 2 - 4 + 1 = -1
\]
Vertex: \((1, -1)\).

3. **Axis of symmetry**: \(x = 1\).

4. **Find the y-intercept**: When \(x = 0\),
\[
y = 1
\]
Y-intercept: \((0, 1)\).

5. **Find x-intercepts**:
\[
2x^2 - 4x + 1 = 0
\]
Use the quadratic formula:
\[
x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} = \frac{4 \pm \sqrt{16 - 8}}{4} = \frac{4 \pm \sqrt{8}}{4} = \frac{4 \pm 2\sqrt{2}}{4} = 1 \pm \frac{\sqrt{2}}{2}
\]
The x-intercepts are:
\[
\left(1 + \frac{\sqrt{2}}{2}, 0\right) \quad \text{and} \quad \left(1 - \frac{\sqrt{2}}{2}, 0\right)
\]

6. **Plot points**: Plot the vertex (1, -1), y-intercept (0, 1), and the x-intercepts calculated above.

7. **Draw the parabola**: Sketch the curve through these points, ensuring it opens upwards since \(a = 2 > 0\).

With practice, graphing quadratic equations becomes more intuitive!