Write a quadratic function using the table given.

For the graph of $y=-2x^2+12x–7$ , write the equation in vertex form.

•  $y=-2\left(x+3\right)^2-61$  

•  $y=-2\left(x+\frac{1}{3}\right)^2+\frac{101}{9}$  

•  $y=-2\left(x-\frac{1}{3}\right)^2-\frac{29}{9}$  

•  $y=-2\left(x-3\right)^2+11$  

Write the equation of a parabola with x-intercepts at (1,0) and (-3,0) and which passes through the point (-2,-9).

Which one is the vertex form of quadratic functions?

• 
  

  $y\ =\left(x\ -\ h\right)^2\ +\ k$
  

• $y\ =\ a\left(x\right)^2+k$

• $y\ \ =\ a\left(x-h\right)^2+k$

• $y\ =\ ax^2+bx+c$

Which is the vertex form of a quadratic equation?

• y= ax²+bc+c

• y= mx + b

• (y-y₁)= m(x-x₁)

• y= a(x-h)²+k

Write the equation of the quadratic if it goes through (-1, 2) and has the given intercepts.

• y=5/2 (x-0)(x-4)
• y = 2/5 (x-0)(x-4)
• y= 2/3 (x-0)(x-4)
• y= 3/2 (x-0)(x-4)

• a = -3, b = -2, c = 5

• a = 3, b = -2, c =-5

• a = -3, b = -2, c =-5

• a = 3, b = 2, c = -5

Is y = 0(x-4)² +2 a quadratic function?

• Not sure, we need more information to determine that.

• Yes, this is in vertex form.

• No, a must be >0 or <0

Write the equation of a parabola with vertex (1, -2), passing through point (3, 10) 

• y= 3(x-1)²+2
• y= 3(x+1)²-2
• y= -3(x-1)²-2
• y= 3(x-1)²-2

Convert the equation y= x²-8x+11 into vertex form.

• y= (x+4)²+5

• y= (x+4)²-5

• y=(x-4)²-5

• y= (x-8)²-53