Quadratic Functions Math Example 3

Follow the full solution, then compare it with the other examples linked below.

Example 3

medium
Find the vertex and axis of symmetry of f(x)=2x2βˆ’8x+3f(x) = 2x^2 - 8x + 3.

Solution

  1. 1
    Identify a=2a = 2, b=βˆ’8b = -8, c=3c = 3. The axis of symmetry is x=βˆ’b2a=βˆ’βˆ’82(2)=84=2x = -\frac{b}{2a} = -\frac{-8}{2(2)} = \frac{8}{4} = 2.
  2. 2
    Find the yy-coordinate of the vertex by evaluating f(2)=2(2)2βˆ’8(2)+3=8βˆ’16+3=βˆ’5f(2) = 2(2)^2 - 8(2) + 3 = 8 - 16 + 3 = -5.
  3. 3
    The vertex is (2,βˆ’5)(2, -5) and the axis of symmetry is the vertical line x=2x = 2. Since a=2>0a = 2 > 0, the parabola opens upward, so the vertex is a minimum.

Answer

Vertex:Β (2,βˆ’5);AxisΒ ofΒ symmetry:Β x=2\text{Vertex: } (2, -5); \quad \text{Axis of symmetry: } x = 2
For f(x)=ax2+bx+cf(x) = ax^2 + bx + c, the vertex occurs at x=βˆ’b2ax = -\frac{b}{2a}. Substituting back gives the yy-coordinate. The axis of symmetry is the vertical line through the vertex. This method avoids completing the square.

About Quadratic Functions

A quadratic function is a polynomial function of degree 2, written as f(x)=ax2+bx+cf(x) = ax^2 + bx + c with a≠0a \neq 0, whose graph is a U-shaped curve called a parabola that opens upward when a>0a > 0 or downward when a<0a < 0.

Learn more about Quadratic Functions β†’

More Quadratic Functions Examples

Example 1 easy

Find the vertex of [formula].

Example 2 medium

Does the parabola [formula] open upward or downward? Find its maximum value.

Example 4 easy

Find the [formula]-intercept of [formula].

Example 5 medium

Find the zeros of [formula].

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