Vertex+Form+Project+-+Olivia,+Sachita,+Pete,+Joe

= //**Vertex Form Project**// =



**a.) Equation in vertex form**: 4(x-1)^2+7  Step 1. y=4x^2-8x+11  Step 2. y=4(x-1)^2+11-4  Step 3. y=4(x-1)^2+7

**b.) Graph Opens:** Up  The graph would open up because the "a" value in the equation is positive. Seen in standard form, (the 4x^2), 4 is positive. In vertex  form of a(x-h)^2+k, the "a" also determines if it opens up or down. The "a" is positive 4, so our parabola would open up.

**c.) Vertex of Equation:** (1,7)  Vertex form is y=a(x-h)^2+k where the vertex is (h,k).  y= 4(x-1)2+7  Since -h is -1, then h is -(-1) which is positive 1 and k is 7. Therefore, the vertex (in this case a minimum) would be (1,7)

**d.) Zero's of Equation**: None (two imaginary numbers)  0=4(x-1)^2+7  -7=4(x-1)^2  √(-7) = 4(x-1)

√(-7)/4 = x-1 ---> [ √(-7)/4 ] +1 = x (which equals (7i/4)+1 ) and [ -√(-7)/4 ] +1 = x which is (-7i/4)+1


 * two imaginary numbers!

**e.) Y-Intercept of Equation**: (0,11)  y-int: (a*h^2)+k  a=4, h=1, k=7  ..................................................  (4*1^2)+7 equals  (4*1)+7 which equals  4+7 = 11

so y-intercept = (0,11)

**f.) Another Quadratic Function with same zeros written in Vertex Form**: 5x^2+2x+2

since our parabola does not have any zeroes, any other quadratic equation can be written that does not have zeroes. To make sure this does not have any zeroes, the discriminant (square root part of the quadratic formula), must be negative. It would be [√(2^2)-(4*5*2)] [√4-40] [√-36] which is a non existing number, meaning that this graph also has no zeroes like the original equation

**g.) Graph of the Equation:**