Question

1. match each equation with its corresponding graph. explain your reasoning.

a) $y = (x - 3)^2$
b) $y = -(x + 4)^2 - 2$
c) $y = -x^2 - 3$
d) $y = (x - 4)^2 + 2$
i) graph with x from -4 to 4, y from -10 to 0, parabola opening down, vertex near (0, -3)
ii) graph with x from 0 to 6, y from 0 to 10, parabola opening up, vertex near (4, 2)
iii) graph with x from -6 to 0, y from -10 to 0, parabola opening down, vertex near (-4, -2)
iv) graph with x from 0 to 6, y from 0 to 10, parabola opening up, vertex near (3, 0)

Explanation:

Step1: Recall Vertex Form of Parabola

The vertex form of a parabola is \( y = a(x - h)^2 + k \), where \((h, k)\) is the vertex. If \( a>0 \), the parabola opens upward; if \( a<0 \), it opens downward.

Step2: Analyze Equation a) \( y=(x - 3)^2 \)

• Here, \( a = 1>0 \) (opens upward), vertex \((h,k)=(3,0)\).
• Look for a parabola opening upward with vertex at \((3,0)\). Graph iv) has vertex at \((3,0)\) and opens upward. So a) matches iv).

Step3: Analyze Equation b) \( y=-(x + 4)^2 - 2 \)

• \( a=-1<0 \) (opens downward), vertex \((h,k)=(-4,-2)\).
• Graph iii) has vertex around \((-4,-2)\) and opens downward. So b) matches iii).

Step4: Analyze Equation c) \( y=-x^2 - 3 \)

• \( a=-1<0 \) (opens downward), vertex \((h,k)=(0,-3)\) (since \( y = - (x - 0)^2 - 3 \)).
• Graph i) has vertex at \((0,-3)\) (approx) and opens downward. So c) matches i).

Step5: Analyze Equation d) \( y=(x - 4)^2 + 2 \)

• \( a = 1>0 \) (opens upward), vertex \((h,k)=(4,2)\).
• Graph ii) has vertex at \((4,2)\) (approx) and opens upward. So d) matches ii).

Answer:

a) matches iv), b) matches iii), c) matches i), d) matches ii)