Question

1. $y=(x + 3)^2$ 24. $y=-(x + 3)^2-4$ 25. $y=-0.5(x + 1)^2+4$ vertex? vertex? vertex? go topic: features of parabolas use the table to identify the vertex, the equation for the axis of symmetry (aos), and state the number of x - intercept(s) the parabola will have, if any. state whether the vertex will be a minimum or a maximum. 26. $\begin{array}{|c|c|}hline x&y\hline - 4&10\hline - 3&3\hline - 2&-2\hline - 1&-5\hline0&-6\hline1&-5\hline2&-2\hlineend{array}$ a. vertex: __ b. aos: c. x - int(s): d. min or max 27. $\begin{array}{|c|c|}hline x&y\hline - 2&49\hline - 1&28\hline0&13\hline1&4\hline2&1\hline3&4\hline4&13\hlineend{array}$ a. vertex: b. aos: c. x - int(s): d. min or max 28. $\begin{array}{|c|c|}hline x&y\hline - 7&-9\hline - 6&3\hline - 5&7\hline - 4&3\hline - 3&-9\hline - 2&-29\hline - 1&-57\hlineend{array}$ a. vertex: b. aos: c. x - int(s): d. min or max 29. $\begin{array}{|c|c|}hline x&y\hline - 8&-9\hline - 7&-8\hline - 6&-9\hline - 5&-12\hline - 4&-17\hline - 3&-24\hline - 2&-33\hlineend{array}$ a. vertex: b. aos: c. x - int(s): __ d. min or max

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.

Step2: Find vertex for $y=(x + 3)^2$

For $y=(x + 3)^2=(x-(-3))^2+0$, the vertex is $(-3,0)$.

Step3: Find vertex for $y=-(x + 3)^2-4$

For $y=-(x + 3)^2-4=-(x-(-3))^2-4$, the vertex is $(-3,-4)$.

Step4: Find vertex for $y=-0.5(x + 1)^2+4$

For $y=-0.5(x + 1)^2+4=-0.5(x-(-1))^2+4$, the vertex is $(-1,4)$.

Step5: Analyze table - based problems

For a parabola, the vertex is the point where the function reaches a maximum or minimum. The axis of symmetry (AoS) has the equation $x = h$ (where $(h,k)$ is the vertex). To find x - intercepts, set $y = 0$ and solve for $x$. If $a>0$ in $y=a(x - h)^2+k$, the vertex is a minimum; if $a<0$, the vertex is a maximum.

For problem 26:

• The vertex is the point where the y - value is the minimum. By observing the table, when $x = 0$, $y=-6$. So the vertex is $(0,-6)$.
• The axis of symmetry is $x = 0$.
• Set $y = 0$. We need to solve the quadratic equation. Since the parabola opens upward (the y - values increase on both sides of $x = 0$), and $y=-6$ at the vertex, there are 2 x - intercepts.
• Since $a>0$ (the parabola opens upward), the vertex is a minimum.

For problem 27:

• The vertex is the point where the y - value is the minimum. By observing the table, when $x = 2$, $y = 1$. So the vertex is $(2,1)$.
• The axis of symmetry is $x = 2$.
• Set $y = 0$. Since the parabola opens upward (the y - values increase on both sides of $x = 2$), and $y = 1$ at the vertex, there are 0 x - intercepts.
• Since $a>0$ (the parabola opens upward), the vertex is a minimum.

For problem 28:

• The vertex is the point where the y - value is the maximum. By observing the table, when $x=-5$, $y = 7$. So the vertex is $(-5,7)$.
• The axis of symmetry is $x=-5$.
• Set $y = 0$. Since the parabola opens downward (the y - values decrease on both sides of $x=-5$), there are 2 x - intercepts.
• Since $a<0$ (the parabola opens downward), the vertex is a maximum.

For problem 29:

• The vertex is the point where the y - value is the maximum. By observing the table, when $x=-7$, $y=-8$. So the vertex is $(-7,-8)$.
• The axis of symmetry is $x=-7$.
• Set $y = 0$. Since the parabola opens downward (the y - values decrease on both sides of $x=-7$), there are 2 x - intercepts.
• Since $a<0$ (the parabola opens downward), the vertex is a maximum.

Answer:

1. Vertex: $(-3,0)$
2. Vertex: $(-3,-4)$
3. Vertex: $(-1,4)$

26.
a. Vertex: $(0,-6)$
b. AoS: $x = 0$
c. x - int(s): 2
d. MIN
27.
a. Vertex: $(2,1)$
b. AoS: $x = 2$
c. x - int(s): 0
d. MIN
28.
a. Vertex: $(-5,7)$
b. AoS: $x=-5$
c. x - int(s): 2
d. MAX
29.
a. Vertex: $(-7,-8)$
b. AoS: $x=-7$
c. x - int(s): 2
d. MAX