which equation would have real zero(s) corresponding to the $x$-intercept(s) of the graph below?

$y = -2^x + 4$
$y = 2^x$
$y = \log_2(x + 2) + 1$
$y = \log_2(-x + 2)$

Question

which equation would have real zero(s) corresponding to the $x$-intercept(s) of the graph below?

$y = -2^x + 4$
$y = 2^x$
$y = \log_2(x + 2) + 1$
$y = \log_2(-x + 2)$

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Exponential Transformations", "Logarithmic Functions" ], "new_concepts": [], "current_concepts": [ "X-intercepts", "Real Zeros", "Exponential Transformations", "Logarithmic Functions" ] } ## Identify the x-intercept from the graph $$ x = 2 $$ ## Find the real zeros of each given equation $$ \begin{aligned} &\text{For } y = -2^x + 4: \quad 0 = -2^x + 4 \implies 2^x = 4 \implies x = 2\ &\text{For } y = 2^x: \quad 0 = 2^x \implies \text{No real solution}\ &\text{For } y = \log_2(x+2) + 1: \quad 0 = \log_2(x+2) + 1 \implies \log_2(x+2) = -1 \implies x+2 = \frac{1}{2} \implies x = -1.5\ &\text{For } y = \log_2(-x+2): \quad 0 = \log_2(-x+2) \implies -x+2 = 1 \implies x = 1 \end{aligned} $$ ## Match the zero to the x-intercept $$ x = 2 \implies y = -2^x + 4 $$

(A) $y = -2^x + 4$

(B) $y = 2^x$

(C) $y = \log_2(x + 2) + 1$

(D) $y = \log_2(-x + 2)$

{ "subject": "Mathematics", "question_type": "Multiple Choice", "knowledge_point": [ "Mathematics", "Algebra", "Exponential Transformations" ] }

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QUESTION IMAGE

Question

Explanation:

Identify the x-intercept from the graph

Find the real zeros of each given equation

Match the zero to the x-intercept

Answer: