Question

on this slide you will need to fill out the table and graph out a parabola on the coordinate plane provided. any work you need to do can be done below in the space provided.
graph the quadratic equation and identify the solution(s).

1. $y = x^2 - 6x + 9$

solutions:

1. $y = x^2 + 4x + 9$

solutions:

1. $y = 2x^2 - 4x + 3$

solutions:

1. $y = \frac{1}{2}x^2 - 4x + 9$

solutions:

1. $y = 2x^2 - 12x + 19$

solutions:

1. $y = -2x^2 - 5x - 10$

solutions:
do your work here. provide the number for the problem above so i can identify which work corresponds to which problem:

Explanation:

Problem 8: $y = x^2 - 6x + 9$

Step1: Pick x-values, compute y

Choose $x=1,2,3,4,5$:

• $x=1$: $y=(1)^2-6(1)+9=1-6+9=4$
• $x=2$: $y=(2)^2-6(2)+9=4-12+9=1$
• $x=3$: $y=(3)^2-6(3)+9=9-18+9=0$
• $x=4$: $y=(4)^2-6(4)+9=16-24+9=1$
• $x=5$: $y=(5)^2-6(5)+9=25-30+9=4$

Step2: Find solution (y=0)

Set $y=0$: $x^2-6x+9=(x-3)^2=0$, so $x=3$.

Table:

x	y
2	1
3	0
4	1
5	4

Step1: Pick x-values, compute y

Choose $x=-4,-3,-2,-1,0$:

• $x=-4$: $y=(-4)^2+4(-4)+9=16-16+9=9$
• $x=-3$: $y=(-3)^2+4(-3)+9=9-12+9=6$
• $x=-2$: $y=(-2)^2+4(-2)+9=4-8+9=5$
• $x=-1$: $y=(-1)^2+4(-1)+9=1-4+9=6$
• $x=0$: $y=(0)^2+4(0)+9=9$

Step2: Check for real solutions

Discriminant: $\Delta = 4^2-4(1)(9)=16-36=-20<0$, so no real solutions.

Table:

x	y
-3	6
-2	5
-1	6
0	9

Step1: Pick x-values, compute y

Choose $x=0,1,2,3,4$:

• $x=0$: $y=2(0)^2-4(0)+3=3$
• $x=1$: $y=2(1)^2-4(1)+3=2-4+3=1$
• $x=2$: $y=2(2)^2-4(2)+3=8-8+3=3$
• $x=3$: $y=2(3)^2-4(3)+3=18-12+3=9$
• $x=4$: $y=2(4)^2-4(4)+3=32-16+3=19$

Step2: Check for real solutions

Discriminant: $\Delta = (-4)^2-4(2)(3)=16-24=-8<0$, so no real solutions.

Table:

x	y
1	1
2	3
3	9
4	19

Answer:

Solution: $x=3$

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Problem 9: $y = x^2 + 4x + 9$