Question

lesson 3.4 checkpoint
complete the previous problems, check your solutions, then complete the lesson checkpoint below.
complete the lesson reflection above by circling your current understanding of the learning goal(s).
error analysis

1. describe and correct the error in finding the vertex of the graph of the function.

$y = -(x + 8)^2$
because $h = -8$, the vertex is $(0, -8)$.
match the function with its graph.

1. $y = -(x + 1)^2 - 3$ ______
2. $y = -(x - 1)^2 + 3$ ______
3. $y = \frac{1}{3}(x - 1)^2 + 3$ ______
4. $y = -2(x + 1)^2 - 3$ ______

a.
\

$$\begin{tikzpicture}scale=0.5 \\draw-> (-5,0) -- (5,0) noderight {$x$}; \\draw-> (0,-5) -- (0,5) nodeabove {$y$}; \\foreach \\x in {-4,-3,-2,-1,1,2,3,4} \\draw (\\x,0.1) -- (\\x,-0.1) nodebelow {$\\x$}; \\foreach \\y in {-4,-3,-2,-1,1,2,3,4} \\draw (0.1,\\y) -- (-0.1,\\y) nodeleft {$\\y$}; \\drawthick, blue, smooth, domain=-4:4 plot (\\x, {- (\\x)^2 + 3}); \\end{tikzpicture}$$

b.
\

$$\begin{tikzpicture}scale=0.5 \\draw-> (-5,0) -- (5,0) noderight {$x$}; \\draw-> (0,-5) -- (0,5) nodeabove {$y$}; \\foreach \\x in {-4,-3,-2,-1,1,2,3,4} \\draw (\\x,0.1) -- (\\x,-0.1) nodebelow {$\\x$}; \\foreach \\y in {-4,-3,-2,-1,1,2,3,4} \\draw (0.1,\\y) -- (-0.1,\\y) nodeleft {$\\y$}; \\drawthick, blue, smooth, domain=-4:4 plot (\\x, { (\\x + 1)^2 - 1}); \\end{tikzpicture}$$

c.
\

$$\begin{tikzpicture}scale=0.5 \\draw-> (-5,0) -- (5,0) noderight {$x$}; \\draw-> (0,-5) -- (0,5) nodeabove {$y$}; \\foreach \\x in {-4,-3,-2,-1,1,2,3,4} \\draw (\\x,0.1) -- (\\x,-0.1) nodebelow {$\\x$}; \\foreach \\y in {-4,-3,-2,-1,1,2,3,4} \\draw (0.1,\\y) -- (-0.1,\\y) nodeleft {$\\y$}; \\drawthick, blue, smooth, domain=-4:4 plot (\\x, { - (\\x + 2)^2 - 3}); \\end{tikzpicture}$$

d.
\

$$\begin{tikzpicture}scale=0.5 \\draw-> (-5,0) -- (5,0) noderight {$x$}; \\draw-> (0,-5) -- (0,5) nodeabove {$y$}; \\foreach \\x in {-4,-3,-2,-1,1,2,3,4} \\draw (\\x,0.1) -- (\\x,-0.1) nodebelow {$\\x$}; \\foreach \\y in {-4,-3,-2,-1,1,2,3,4} \\draw (0.1,\\y) -- (-0.1,\\y) nodeleft {$\\y$}; \\drawthick, blue, smooth, domain=-4:4 plot (\\x, { (\\x)^2 + 1}); \\end{tikzpicture}$$

Explanation:

Step1: Recall vertex form rules

The vertex form of a parabola is $y=a(x-h)^2+k$, where:

• $(h,k)$ is the vertex
• $a>0$ opens upward, $a<0$ opens downward
• $|a|>1$ narrows, $|a|<1$ widens the parabola

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Problem 30: Error Analysis

Step1: Identify vertex form mismatch

Given $y=-(x+8)^2 = -(x-(-8))^2$, so $h=-8, k=0$.

Step2: Correct vertex calculation

Vertex is $(h,k)=(-8, 0)$. The original error swapped $h$ and $k$ values incorrectly.

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Problem 35: $y=-(x+1)^2-3$

Step1: Find vertex and direction

Rewrite as $y=-(x-(-1))^2+(-3)$: vertex $(-1,-3)$, $a=-1<0$ (opens downward).

Step2: Match to graph

This matches Graph C.

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Problem 36: $y=-(x-1)^2+3$

Step1: Find vertex and direction

Vertex $(1,3)$, $a=-1<0$ (opens downward).

Step2: Match to graph

This matches Graph A.

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Problem 37: $y=\frac{1}{3}(x-1)^2+3$

Step1: Find vertex and direction

Vertex $(1,3)$, $a=\frac{1}{3}>0$ (opens upward, wide).

Step2: Match to graph

This matches Graph D.

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Problem 38: $y=-2(x+1)^2-3$

Step1: Find vertex and direction

Rewrite as $y=-2(x-(-1))^2+(-3)$: vertex $(-1,-3)$, $a=-2<0$ (opens downward, narrow).

Step2: Match to graph

This matches Graph B.

Answer:

1. Error: Incorrectly identified the vertex as $(0,-8)$ instead of using the vertex form rule. Correction: For $y=-(x+8)^2=-(x-(-8))^2$, the vertex is $(-8, 0)$.
2. C
3. A
4. D
5. B