which function has a vertex on the $y$-axis?

$f(x) = (x - 2)^2$
$f(x) = x(x + 2)$
$f(x) = (x - 2)(x + 2)$
$f(x) = (x + 1)(x - 2)$

Question

which function has a vertex on the $y$-axis?

$f(x) = (x - 2)^2$
$f(x) = x(x + 2)$
$f(x) = (x - 2)(x + 2)$
$f(x) = (x + 1)(x - 2)$

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": false, "concepts_used": [ "Quadratic Vertex", "Intercept Form" ], "new_concepts": [], "current_concepts": [ "Quadratic Vertex", "Intercept Form" ] } ## Define condition for vertex on y-axis Using the *Quadratic Vertex* knowledge point $$ x_v = 0 $$ ## Analyze the first function Using the *Quadratic Vertex* knowledge point $$ \begin{aligned} &f(x) = (x-2)^2\ &x_v = 2 e 0 \end{aligned} $$ ## Analyze the second function Using the *Intercept Form* and *Quadratic Vertex* knowledge points $$ \begin{aligned} &f(x) = x(x+2)\ &x_1 = 0,\quad x_2 = -2\ &x_v = \frac{0 + (-2)}{2} = -1 e 0 \end{aligned} $$ ## Analyze the third function Using the *Intercept Form* and *Quadratic Vertex* knowledge points $$ \begin{aligned} &f(x) = (x-2)(x+2)\ &x_1 = 2,\quad x_2 = -2\ &x_v = \frac{2 + (-2)}{2} = 0 \end{aligned} $$ ## Analyze the fourth function Using the *Intercept Form* and *Quadratic Vertex* knowledge points $$ \begin{aligned} &f(x) = (x+1)(x-2)\ &x_1 = -1,\quad x_2 = 2\ &x_v = \frac{-1 + 2}{2} = 0.5 e 0 \end{aligned} $$

(A) $f(x) = (x - 2)^2$

(B) $f(x) = x(x + 2)$

(C) $f(x) = (x - 2)(x + 2)$

(D) $f(x) = (x + 1)(x - 2)$

{ "subject": "Mathematics", "question_type": "Multiple Choice", "knowledge_point": [ "Mathematics", "Algebra", "Quadratic Vertex" ] }

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QUESTION IMAGE

Question

Explanation:

Define condition for vertex on y-axis

Analyze the first function

Analyze the second function

Analyze the third function

Analyze the fourth function

Answer: