Vertex Form: Axis of Symmetry, Zeros, and Example g(x)=3(x-2)^2-5

Vertex Form Overview

  • Vertex form of a quadratic: g(x)=a(x−h)2+kg(x)=a(x-h)^2+k
    • Vertex is at the point $(h,k)$.
    • Axis of symmetry is the vertical line x=hx=h.
    • The parabola opens up if a>0 and down if a<0.
  • In the example from the transcript:
    • The quadratic is written in vertex form as g(x)=3(x−2)2−5g(x)=3(x-2)^2-5.
    • Here a=3a=3, h=2h=2, k=−5k=-5, so the vertex is (2,−5)(2,-5) and the axis of symmetry is x=2x=2.
    • This matches the intuition: the graph is symmetric about the vertical line x=2x=2.

Example evaluations illustrate symmetry

  • Evaluate at several x-values:
    • g(0)=3(0−2)2−5=3 ⋅ 4−5=12−5=7g(0)=3(0-2)^2-5=3\,\cdot\,4-5=12-5=7
    • g(4)=3(4−2)2−5=3 ⋅ 4−5=7g(4)=3(4-2)^2-5=3\,\cdot\,4-5=7
    • g(1)=3(1−2)2−5=3 ⋅ 1−5=−2g(1)=3(1-2)^2-5=3\,\cdot\,1-5=-2
  • Observations:
    • The values at x=0 and x=4 are the same because they are the same distance from the axis $x=2$.
    • Distances from the axis: ∣0−2∣=2|0-2|=2 and ∣4−2∣=2|4-2|=2; hence the same output due to the squared term.
    • In general, for any d<br/>eq0d<br /> eq 0, g(2+d)=g(2−d)g(2+d)=g(2-d) because (2+d−2)2=(2−d−2)2=d2<br/>(2+d-2)^2=(2-d-2)^2=d^2 <br />.
  • This symmetry is what makes x=2x=2 the axis of symmetry and explains the same results on either side of the axis.

Axis of symmetry and vertex form connection

  • The line x=hx=h is the axis of symmetry for any quadratic in vertex form g(x)=a(x−h)2+kg(x)=a(x-h)^2+k.
  • The vertex form generalizes to any quadratic in the form a(x−h)2+ka(x-h)^2+k.
  • The axis of symmetry is a direct consequence of the squared term: the parabola is mirrored on either side of x=hx=h.

Zeros (x-intercepts) from vertex form

  • To find zeros of g(x)=a(x−h)2+k",setg(x)=a(x-h)^2+k", setg(x)=0":
    • a(x−h)2+k=0⇒(x−h)2=−ka.a(x-h)^2+k=0\quad\Rightarrow\quad (x-h)^2=-\frac{k}{a}.
    • If −ka≥0-\frac{k}{a} \ge 0, then the zeros are x=h±−ka.x=h\pm\sqrt{-\frac{k}{a}}.
  • For the example g(x)=3(x−2)2−5g(x)=3(x-2)^2-5:
    • Set to zero: 3(x−2)2−5=0⇒(x−2)2=53.3(x-2)^2-5=0\Rightarrow (x-2)^2=\frac{5}{3}.
    • Zeros: x=2±53=2±153.x=2\pm\sqrt{\frac{5}{3}} = 2\pm\frac{\sqrt{15}}{3}.
    • Approximate zeros: x≈2−1.290=0.710andx≈2+1.290=3.290.x\approx 2-1.290=0.710\quad\text{and}\quad x\approx 2+1.290=3.290.
  • Key takeaway:
    • Zeros are symmetric about the axis of symmetry, so the two zeros are equally distant from x=hx=h.

Quick summary and connections

  • Vertex form lets you read off:
    • Vertex: (h,k)(h,k)
    • Axis of symmetry: x=hx=h
    • Zeros by solving a(x−h)2+k=0a(x-h)^2+k=0, yielding x=h±−kax=h\pm\sqrt{-\frac{k}{a}} when real.
  • In real-world terms, the axis of symmetry is the line that divides the parabola into two mirror-image halves.
  • The same-distance property around the axis explains why values at equidistant x-values from the axis are equal.
  • This approach generalizes to any quadratic expressed in vertex form, making it a powerful tool for graphing and solving quadratics.