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Unit 6: Quadratic and Polynomial Functions

Topic 4

Graphing Parabolas (Vertex and Intercepts)

A parabola is the graph of a quadratic function:
y=ax2+bx+c.y = ax^2 + bx + c.
It is a U-shaped curve that opens upward when a>0a > 0 and downward when a<0.a < 0.
Key features of a parabola include its:
  • Vertex: the highest or lowest point.
  • Axis of symmetry: the vertical line that passes through the vertex.
  • Y-intercept: where the graph crosses the y-axis.
  • X-intercepts (zeros): where the graph crosses the x-axis, found by solving ax2+bx+c=0.ax^2 + bx + c = 0.
For y=x2−4x+3y = x^2 - 4x + 3, these features look like this:
The vertex can be found from the formula:
x=−b2a,y=f ⁣(−b2a).x = -\frac{b}{2a}, \quad y = f\!\left(-\frac{b}{2a}\right).
In vertex form, y=a(x−h)2+k,y = a(x - h)^2 + k, the vertex is (h,k)(h, k). The axis of symmetry is x=h.x = h.
Why It Works Completing the square converts y=ax2+bx+cy = ax^2 + bx + c into vertex form, revealing how the graph shifts horizontally and vertically from the parent function y=x2.y = x^2.
(0, 3)(1, 0)(3, 0)Vertex (2, -1)

Core Skills

  • Identify vertex, axis of symmetry, and intercepts from any form.
  • Convert between standard and vertex forms.
  • Determine whether the parabola opens up or down.
  • Sketch parabolas accurately by plotting key points.
  • Interpret graphs in real-world contexts (maximums/minimums).

Example 1: Finding Vertex and Intercepts

Graph y=x2−4x+3.y = x^2 - 4x + 3. Step 1: Find the vertex.
x=−b2a=42=2,y=(2)2−4(2)+3=4−8+3=−1.x = -\frac{b}{2a} = \frac{4}{2} = 2, \quad y = (2)^2 - 4(2) + 3 = 4 - 8 + 3 = -1.
Vertex: (2,−1)(2, -1). Step 2: Find the intercepts. Y-intercept: when x=0,  y=3.x = 0,\; y = 3. X-intercepts: solve x2−4x+3=0⇒(x−3)(x−1)=0⇒x=1,3.x^2 - 4x + 3 = 0 \Rightarrow (x - 3)(x - 1) = 0 \Rightarrow x = 1, 3. Step 3: Direction: a=1>0a = 1 > 0, so the parabola opens upward.

Example 2: Vertex Form and Transformations

Graph y=2(x+1)2−3.y = 2(x + 1)^2 - 3. Vertex: (−1,−3)(-1, -3). Axis of symmetry: x=−1.x = -1. Opening: Upward (a=2>0a = 2 > 0). Stretch: Narrower than y=x2.y = x^2.

Example 3: Downward Opening Parabola

Graph y=−x2+2x+3.y = -x^2 + 2x + 3. Step 1: Vertex: x=−b2a=−2−2=1.x = -\frac{b}{2a} = -\frac{2}{-2} = 1. y=−(1)2+2(1)+3=−1+2+3=4.y = -(1)^2 + 2(1) + 3 = -1 + 2 + 3 = 4. Vertex: (1,4)(1, 4). Step 2: Intercepts: Y-intercept: when x=0,  y=3.x = 0,\; y = 3. X-intercepts: solve −x2+2x+3=0-x^2 + 2x + 3 = 0. Multiply by –1: x2−2x−3=0⇒(x−3)(x+1)=0.x^2 - 2x - 3 = 0 \Rightarrow (x - 3)(x + 1) = 0. x=3,−1\boxed{x = 3, -1}.

Example 4: Finding the Equation from a Graph

A parabola has vertex (2,−4)(2, -4) and passes through the point (0,4)(0, 4). Find its equation. Use vertex form y=a(x−h)2+k.y = a(x - h)^2 + k.
y=a(x−2)2−4.y = a(x - 2)^2 - 4.
Substitute (x,y)=(0,4):(x, y) = (0, 4):
4=a(0−2)2−4⇒4=4a−4⇒8=4a⇒a=2.4 = a(0 - 2)^2 - 4 \Rightarrow 4 = 4a - 4 \Rightarrow 8 = 4a \Rightarrow a = 2.
Equation: y=2(x−2)2−4.\boxed{y = 2(x - 2)^2 - 4.}

Example 5: Real-World Interpretation

A rocket’s height is given by h=−16t2+64t+80.h = -16t^2 + 64t + 80. Find its vertex and interpret.
t=−b2a=−64−32=2.t = -\frac{b}{2a} = -\frac{64}{-32} = 2.
h(2)=−16(4)+64(2)+80=−64+128+80=144.h(2) = -16(4) + 64(2) + 80 = -64 + 128 + 80 = 144. Vertex: (2,144)(2, 144). Interpretation: The rocket reaches a maximum height of 144 ft at 2 seconds.

Key Takeaways

  • Use x=−b2ax = -\frac{b}{2a} to find the vertex in standard form.
  • The sign of aa determines if the parabola opens up or down.
  • The axis of symmetry passes through the vertex.
  • The vertex represents a minimum if a>0a > 0, or a maximum if a<0.a < 0.
  • Plot vertex, intercepts, and symmetry to sketch accurate graphs.