if (a 0), which of the following could be an equation of the graph shown in the figure to the right?

(a) (y = ax^2)
(b) (y = x^2 - a)
(c) (y = x^2 + a)
(d) (y = (x - a)^2)
(e) (y = (x + a)^2)

Question

if (a > 0), which of the following could be an equation of the graph shown in the figure to the right?

(a) (y = ax^2)
(b) (y = x^2 - a)
(c) (y = x^2 + a)
(d) (y = (x - a)^2)
(e) (y = (x + a)^2)

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [], "new_concepts": [ "Quadratic Graphs", "Vertical Translations" ], "current_concepts": [ "Quadratic Graphs", "Vertical Translations", "Parabola Vertex" ] } ## Analyze the graph features $$ \begin{aligned} &\text{Vertex is on the } y\text{-axis above the origin: } (0, k) \text{ where } k > 0\ &\text{Parabola opens upward: } y = x^2 + k \end{aligned} $$ ## Evaluate the options given $a > 0$ $$ \begin{aligned} &\text{(A) } y = ax^2 \implies \text{Vertex is at } (0,0)\ &\text{(B) } y = x^2 - a \implies \text{Vertex is at } (0, -a) \text{ with } -a < 0\ &\text{(C) } y = x^2 + a \implies \text{Vertex is at } (0, a) \text{ with } a > 0\ &\text{(D) } y = (x - a)^2 \implies \text{Vertex is at } (a, 0) \text{ on the positive } x\text{-axis}\ &\text{(E) } y = (x + a)^2 \implies \text{Vertex is at } (-a, 0) \text{ on the negative } x\text{-axis} \end{aligned} $$ ## Match graph with correct option $$ \text{The graph vertex is at } (0, k) \text{ with } k > 0 \implies y = x^2 + a $$

(A) $y = ax^2$

(B) $y = x^2 - a$

(C) $y = x^2 + a$

(D) $y = (x - a)^2$

(E) $y = (x + a)^2$

{ "subject": "Mathematics", "question_type": "Multiple Choice", "knowledge_point": [ "Mathematics", "Geometry", "Quadratic Graphs" ] }

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QUESTION IMAGE

Question

Explanation:

Analyze the graph features

Evaluate the options given \(a > 0\)

Match graph with correct option

Answer: