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Sagot :

Remember the shortcut way for graphing quadratic equations

  • A quadratic function has graph as parabola
  • Hence on both sides of vertex the parabola is symmetric and axis of symmetry is vertex x values .
  • y on both sides for -x and +x is same

#1

  • y=2(x-3)²-1

Vertex

  • (h,k)=(3,-1)

As a is positive parabola facing upwards

Find y for same x distance from vertex

I took 3-1=2 and 3+1=4

  • f(2)=2(2-3)²-1=2(-1)²-1=2-1=1
  • f(4)=2(4-3)²-1=1

Now plot vertex and these two points (2,1) and (4,1) on graph then draw a parabola by freehand

#2

  • y=(x-2)(x+4)
  • y=x²+4x-2x-8
  • y=x²+2x-8

Convert to vertex form

  • y=x²+2x+1-9
  • y=(x+1)²-9

Vertex at (-1,-9)

Same take two equidistant x values

Let's take -1-1=-2 and -1+1 =0

  • f(-2)=(-2+1)²-9=1-9=-8
  • f(0)=(1)²-9=-8

Put (-1,-9),(-2,-8),(0,-8) on graph and draw a freehand parabola

#3.

Yes it can be verified by finding the coordinate theoretically on putting them on function then can be verified through putting them on graph whether they matches or not

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