T1W5L3

#methods#ae#lesson

quadratic equations §

• many situations can be modelled using quadratic equation, so it is useful to be able to solve them.
  1. rearrange the equation into the form $a{x}^2+bx+c=0$
     1. you may need to expand and simplify first.
  2. factorise the quadratic, using the techniques.
  3. use null factor theorem (if ab = 0, then a = 0 or b = 0) to solve for x

practice: solve $2x^2+5x-12=0$

• factors of $2\times -12$ which make $5$ is $-3$ and $-8$
• $\frac{(2x-3)(2x+8)}{2}=0$
• $\frac{(2x-3)(x+4)\times 2}{2}=0$
• $(2x-3)(x+4)=0$ OR
• $2x^2-3x+8x-12=0$
• $x(2x-3)+4(2x-3)=0$
• $2x-3=0$ or $x+4=0$
• $x=\frac{3}{2},-4$

quadratics in turning point form §

• quadratics in turning point form $y=a(x-h{)}^2+k$ can be visualised as transformations of $y=x^2$

• https://www.geogebra.org/m/EFbtkvVP

• when a is positive:

  • parabola is concave up
  • the turning point is a minimum

• when a is negative, the parabola reflects about the x-axis

  • the parabola is concave down
  • the turning point is a maximum

• the parabola dilates from the x-axis by a factor of a

  • when $|a|>1$, the parabola is ‘stretched’ vertically.
  • when $|a|<1$, the parabola is squashed vertically.
  • the dilations are NOT horizontal !!

• $y=a{x}^2$ to $y=a(x-h{)}^2+k$

• the parabola

  • translates right $h$ units
    • left if $h$ is negative
  • translates up $k$ units
    • down if $k$ is negative
  • the turning point is at ($h$,$k$)
  • the axis of symmetry is at $x=h$

• for the equation $y=-(x+1{)}^2+4$

• a. axis of symmetry

  • x = -1

• b. turning point and nature

  • (-1, 4) MAX

• c. y intercept

  • $y=-(0+1{)}^2+4$
  • $=-1+4=3$
  • (0, 3)

• d. x intercept

  • $0=-(x+1{)}^2+4$
  • $4=(x+1{)}^2$
  • $\pm 2=x+1$
  • $x+1=2$ or $x+1=-2$
  • $x=3$ or $x=-1$
  • $(1,0)$ or $(-3,0)$