Graphing Quadratic Functions

Parabolas, vertex, axis of symmetry

Graphing Quadratic Functions

Standard Form

$y = ax^2 + bx + c$

The graph is a parabola.

Direction of Opening

If $a > 0$ : parabola opens upward (U-shaped)
If $a < 0$ : parabola opens downward (∩-shaped)

Vertex

The vertex is the highest or lowest point on the parabola.

Vertex formula: $x = -\frac{b}{2a}$

Then substitute to find $y$ -coordinate.

Axis of Symmetry

A vertical line through the vertex: $x = -\frac{b}{2a}$

Y-Intercept

The point where the graph crosses the y-axis: $(0, c)$

Vertex Form

$y = a(x - h)^2 + k$

Vertex is at $(h, k)$

📚 Practice Problems

1Problem 1easy

❓ Question:

Does the parabola $y = -2x^2 + 3x + 1$ open upward or downward?

💡 Show Solution

Look at the coefficient of $x^2$ :

$a = -2$

Since $a < 0$ (negative), the parabola opens downward.

Answer: Downward

2Problem 2medium

❓ Question:

Find the vertex of $y = x^2 - 6x + 5$

💡 Show Solution

Identify: $a = 1$ , $b = -6$ , $c = 5$

Step 1: Find the x-coordinate of the vertex $x = -\frac{b}{2a} = -\frac{-6}{2(1)} = \frac{6}{2} = 3$

Step 2: Find the y-coordinate by substituting $x = 3$ $y = (3)^2 - 6(3) + 5$ $y = 9 - 18 + 5$ $y = -4$

Answer: Vertex is at $(3, -4)$

3Problem 3medium

❓ Question:

What is the axis of symmetry for $y = 2x^2 + 8x - 3$ ?

💡 Show Solution

The axis of symmetry is the vertical line through the vertex.

Use: $x = -\frac{b}{2a}$

Identify: $a = 2$ , $b = 8$

$x = -\frac{8}{2(2)} = -\frac{8}{4} = -2$

Answer: $x = -2$

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