Level 36 • 50/50 questions • 40 seconds per question.
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What is the slope of the graph of (f(x)=x)?
A. (1)
Comparing (y=x) with (y=mx+c) gives (m=1). So its slope is (1).
The correct answer is A. (1). Comparing (y=x) with (y=mx+c) gives (m=1). So its slope is (1).
(y=x) को (y=mx+c) से मिलाने पर (m=1) मिलता है। इसलिए इसकी ढाल (1) है।
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What is the (y)-intercept of the graph of (f(x)=5)?
B. (5)
Putting (x=0) gives (y=5). The (y)-intercept of a constant function is the constant value itself.
The correct answer is B. (5). Putting (x=0) gives (y=5). The (y)-intercept of a constant function is the constant value itself.
(x=0) रखने पर (y=5) मिलता है। नियत फलन का (y)-अक्ष प्रतिच्छेद वही नियत मान होता है।
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The line (y=-2) is parallel to which axis?
A. (x)-अक्ष(x)-axis
In (y=-2), (y) is fixed and (x) varies. So it is a horizontal line parallel to the (x)-axis.
The correct answer is A. (x)-अक्ष / (x)-axis. In (y=-2), (y) is fixed and (x) varies. So it is a horizontal line parallel to the (x)-axis.
(y=-2) में (y) स्थिर रहता है और (x) बदलता है। इसलिए यह (x)-अक्ष के समांतर क्षैतिज रेखा है।
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Which point lies on the graph of (f(x)=x-2)?
A. ((3,9))
Since \(3^2=9\), ((3,9)) lies on the graph. To check a point, put (x) in (y=f(x)).
The correct answer is A. ((3,9)). Since \(3^2=9\), ((3,9)) lies on the graph. To check a point, put (x) in (y=f(x)).
\(3^2=9\), इसलिए ((3,9)) आलेख पर स्थित है। बिंदु जांचने के लिए (y=f(x)) में (x) रखें।
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What is the maximum value of (f(x)=-x-2+1)?
B. (1)
Since \(-x^2\le 0\), \(-x^2+1\le 1\). The maximum value (1) occurs at (x=0).
The correct answer is B. (1). Since \(-x^2\le 0\), \(-x^2+1\le 1\). The maximum value (1) occurs at (x=0).
क्योंकि \(-x^2\le 0\), इसलिए \(-x^2+1\le 1\)। अधिकतम मान (1) (x=0) पर मिलता है।
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In which direction is the graph of (f(x)=(x-4)2) shifted from \(y=x^2\)?
A. (4) इकाई दाईं ओर(4) units right
In ((x-4)2), the inside expression is (x-4). So the graph shifts (4) units right.
The correct answer is A. (4) इकाई दाईं ओर / (4) units right. In ((x-4)2), the inside expression is (x-4). So the graph shifts (4) units right.
((x-4)2) में अंदर (x-4) है। इसलिए आलेख (4) इकाई दाईं ओर खिसकता है।
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What is the vertex of (f(x)=(x+3)2+2)?
A. ((-3,2))
((x+3)2) becomes zero at (x=-3) and (2) is added outside. So the vertex is ((-3,2)).
The correct answer is A. ((-3,2)). ((x+3)2) becomes zero at (x=-3) and (2) is added outside. So the vertex is ((-3,2)).
((x+3)2) शून्य (x=-3) पर होता है और बाहर (2) जुड़ता है। इसलिए शीर्ष ((-3,2)) है।
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At which (x)-value does (f(x)=|x-4|) get its minimum value?
B. (x=4)
The minimum value of (|x-4|) is (0) when (x-4=0). So (x=4) is correct.
The correct answer is B. (x=4). The minimum value of (|x-4|) is (0) when (x-4=0). So (x=4) is correct.
(|x-4|) का न्यूनतम मान (0) तब होता है जब (x-4=0)। इसलिए (x=4) सही है।
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How is the graph of (f(x)=|x+5|) changed from (y=|x|)?
A. (5) इकाई बाईं ओर(5) units left
The inside term (x+5) shifts the graph (5) units left. The inside sign shows the opposite direction.
The correct answer is A. (5) इकाई बाईं ओर / (5) units left. The inside term (x+5) shifts the graph (5) units left. The inside sign shows the opposite direction.
अंदर (x+5) होने से ग्राफ (5) इकाई बाईं ओर खिसकता है। अंदर का चिन्ह दिशा उलटी बताता है।
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How is the graph of (f(x)=2|x|) compared with (y=|x|)?
A. ऊर्ध्व रूप से खिंचा हुआVertically stretched
In (2|x|), every (y)-value becomes double. So the graph is vertically stretched.
The correct answer is A. ऊर्ध्व रूप से खिंचा हुआ / Vertically stretched. In (2|x|), every (y)-value becomes double. So the graph is vertically stretched.
(2|x|) में हर (y)-मान दोगुना हो जाता है। इसलिए ग्राफ ऊर्ध्व रूप से खिंचता है।
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What is the starting point of (f(x)=\sqrt{x+4})?
A. ((-4,0))
In \(\sqrt{x+4}\), (x+4=0) gives (x=-4). So the starting point is ((-4,0)).
The correct answer is A. ((-4,0)). In \(\sqrt{x+4}\), (x+4=0) gives (x=-4). So the starting point is ((-4,0)).
\(\sqrt{x+4}\) में (x+4=0) से (x=-4) मिलता है। इसलिए प्रारंभिक बिंदु ((-4,0)) है।
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What is the domain of (f(x)=\sqrt{x-5})?
A. \(x\ge 5\)
For a real square root, \(x-5\ge 0\) is required. Hence the domain is \(x\ge 5\).
The correct answer is A. \(x\ge 5\). For a real square root, \(x-5\ge 0\) is required. Hence the domain is \(x\ge 5\).
वास्तविक वर्गमूल के लिए \(x-5\ge 0\) चाहिए। इसलिए डोमेन \(x\ge 5\) है।
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Which point is not on the graph of (f(x)=\sqrt{x})?
D. ((-1,1))
For \(\sqrt{x}\), \(x\ge 0\) is required. In ((-1,1)), (x=-1), so this point is not on the graph.
The correct answer is D. ((-1,1)). For \(\sqrt{x}\), \(x\ge 0\) is required. In ((-1,1)), (x=-1), so this point is not on the graph.
\(\sqrt{x}\) के लिए \(x\ge 0\) होना चाहिए। ((-1,1)) में (x=-1) है इसलिए यह बिंदु ग्राफ पर नहीं है।
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How is the graph of (f(x)=x-3-2) changed from \(y=x^3\)?
A. (2) इकाई नीचे(2) units down
The outside (-2) decreases every (y)-value by (2). So the graph shifts (2) units down.
The correct answer is A. (2) इकाई नीचे / (2) units down. The outside (-2) decreases every (y)-value by (2). So the graph shifts (2) units down.
बाहर (-2) होने से हर (y)-मान (2) कम हो जाता है। इसलिए ग्राफ (2) इकाई नीचे खिसकता है।
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In which direction is (f(x)=(x+2)3) shifted from \(y=x^3\)?
A. (2) इकाई बाईं ओर(2) units left
The inside term (x+2) shifts the graph (2) units left. In horizontal shifts, the sign is opposite.
The correct answer is A. (2) इकाई बाईं ओर / (2) units left. The inside term (x+2) shifts the graph (2) units left. In horizontal shifts, the sign is opposite.
अंदर (x+2) होने से ग्राफ (2) इकाई बाईं ओर जाता है। क्षैतिज खिसकाव में चिन्ह उल्टा होता है।
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The graph of (f(x)=-x-3) is the reflection of \(y=x^3\) in which axis?
A. (x)-अक्ष(x)-axis
The outside negative sign changes the sign of every (y)-value. So it is a reflection in the (x)-axis.
The correct answer is A. (x)-अक्ष / (x)-axis. The outside negative sign changes the sign of every (y)-value. So it is a reflection in the (x)-axis.
बाहर ऋण चिह्न हर (y)-मान का चिह्न बदलता है। इसलिए यह (x)-अक्ष में प्रतिबिंब है।
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Which (x)-value is not allowed in (f(x)=\frac{1}{x-3})?
A. (x=3)
The denominator (x-3) cannot be zero. From (x-3=0), we get (x=3).
The correct answer is A. (x=3). The denominator (x-3) cannot be zero. From (x-3=0), we get (x=3).
हर (x-3) शून्य नहीं हो सकता। (x-3=0) से (x=3) मिलता है।
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What is the range of (f(x)=\frac{1}{x}+4)?
A. (\(-\infty,4\)\cup\(4,\infty\))
\(\frac{1}{x}\) is never (0), so \(\frac{1}{x}+4\) is never (4). Thus all values except (4) are obtained.
The correct answer is A. (\(-\infty,4\)\cup\(4,\infty\)). \(\frac{1}{x}\) is never (0), so \(\frac{1}{x}+4\) is never (4). Thus all values except (4) are obtained.
\(\frac{1}{x}\) कभी (0) नहीं होता इसलिए \(\frac{1}{x}+4\) कभी (4) नहीं होता। अतः (4) को छोड़कर सभी मान मिलते हैं।
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In which quadrants does the graph of (f(x)=\frac{2}{x}) lie?
A. प्रथम और तृतीयFirst and third
In \(\frac{2}{x}\), (x) and (y) have the same sign. So the graph lies in the first and third quadrants.
The correct answer is A. प्रथम और तृतीय / First and third. In \(\frac{2}{x}\), (x) and (y) have the same sign. So the graph lies in the first and third quadrants.
\(\frac{2}{x}\) में (x) और (y) के चिह्न समान होते हैं। इसलिए ग्राफ प्रथम और तृतीय चतुर्थांश में होता है।
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In which quadrants does the graph of (f(x)=\frac{-3}{x}) lie?
B. द्वितीय और चतुर्थSecond and fourth
In \(\frac{-3}{x}\), (x) and (y) have opposite signs. So the graph lies in the second and fourth quadrants.
The correct answer is B. द्वितीय और चतुर्थ / Second and fourth. In \(\frac{-3}{x}\), (x) and (y) have opposite signs. So the graph lies in the second and fourth quadrants.
\(\frac{-3}{x}\) में (x) और (y) के चिह्न विपरीत होते हैं। इसलिए ग्राफ द्वितीय और चतुर्थ चतुर्थांश में होता है।
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What is the horizontal asymptote of (f(x)=\frac{1}{x-2}+1)?
B. (y=1)
For large (|x|), \(\frac{1}{x^2}\) approaches (0). So the horizontal asymptote is (y=1).
The correct answer is B. (y=1). For large (|x|), \(\frac{1}{x^2}\) approaches (0). So the horizontal asymptote is (y=1).
बड़े (|x|) पर \(\frac{1}{x^2}\) (0) के पास जाता है। इसलिए क्षैतिज आसिम्प्टोट (y=1) है।
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For (f(x)=\operatorname{sgn}(x-2)), what is the value at (x=5)?
A. (1)
At (x=5), (x-2=3>0), so the signum value is (1). First check the sign of the inside expression.
The correct answer is A. (1). At (x=5), (x-2=3>0), so the signum value is (1). First check the sign of the inside expression.
(x=5) पर (x-2=3>0), इसलिए साइनम मान (1) है। पहले अंदर की मात्रा का चिह्न देखें।
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For (f(x)=\operatorname{sgn}(x+1)), what is the value at (x=-1)?
B. (0)
At (x=-1), (x+1=0), so (\operatorname{sgn}(0)=0). In signum, the value at zero is separate.
The correct answer is B. (0). At (x=-1), (x+1=0), so (\operatorname{sgn}(0)=0). In signum, the value at zero is separate.
(x=-1) पर (x+1=0), इसलिए (\operatorname{sgn}(0)=0)। साइनम में शून्य का मान अलग होता है।
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For (f(x)=\operatorname{sgn}(x+4)), what is the value when (x<-4)?
C. (-1)
If (x<-4), then (x+4<0). So the value of the signum function is (-1).
The correct answer is C. (-1). If (x<-4), then (x+4<0). So the value of the signum function is (-1).
यदि (x<-4), तो (x+4<0)। इसलिए साइनम फलन का मान (-1) होगा।
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For (f(x)=\lfloor x\rfloor), what is the value at (x=4)?
B. (4)
When (x) is an integer, \(\lfloor x\rfloor=x\). Therefore \(\lfloor 4\rfloor=4\).
The correct answer is B. (4). When (x) is an integer, \(\lfloor x\rfloor=x\). Therefore \(\lfloor 4\rfloor=4\).
जब (x) पूर्णांक हो तो \(\lfloor x\rfloor=x\) होता है। इसलिए \(\lfloor 4\rfloor=4\)।
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For (f(x)=\lfloor x\rfloor), what is the value at (x=-3)?
B. (-3)
For an integer input, the greatest integer function gives the same number. So \(\lfloor -3\rfloor=-3\).
The correct answer is B. (-3). For an integer input, the greatest integer function gives the same number. So \(\lfloor -3\rfloor=-3\).
पूर्णांक इनपुट पर ग्रेटेस्ट इंटीजर फलन वही संख्या देता है। इसलिए \(\lfloor -3\rfloor=-3\) है।
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For (f(x)=\lfloor x\rfloor), what is the value at (x=0.8)?
A. (0)
\(\lfloor 0.8\rfloor=0\), because (0) is the greatest integer less than or equal to (0.8). For decimals, take the lower integer.
The correct answer is A. (0). \(\lfloor 0.8\rfloor=0\), because (0) is the greatest integer less than or equal to (0.8). For decimals, take the lower integer.
\(\lfloor 0.8\rfloor=0\), क्योंकि (0.8) से छोटी या बराबर सबसे बड़ी पूर्णांक संख्या (0) है। दशमलव में नीचे की पूर्णांक संख्या लें।
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In the graph of (f(x)=\lfloor x\rfloor), what is the value of (y) on ([-1,0))?
A. (-1)
For every \(x\in[-1,0\)), \(\lfloor x\rfloor=-1\). So (y=-1) on this part.
The correct answer is A. (-1). For every \(x\in[-1,0\)), \(\lfloor x\rfloor=-1\). So (y=-1) on this part.
हर \(x\in[-1,0\)) के लिए \(\lfloor x\rfloor=-1\)। इसलिए इस भाग में (y=-1) रहता है।
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For the fractional part function (f(x)={x}), what is the value at (x=2.6)?
A. (0.6)
The fractional part is \(x-\lfloor x\rfloor\). Hence ({2.6}=2.6-2=0.6).
The correct answer is A. (0.6). The fractional part is \(x-\lfloor x\rfloor\). Hence ({2.6}=2.6-2=0.6).
भिन्नात्मक भाग \(x-\lfloor x\rfloor\) होता है। इसलिए ({2.6}=2.6-2=0.6)।
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What is the usual range of the fractional part function (f(x)={x})?
A. ([0,1))
The fractional part is never less than (0) and is always less than (1). So the range is ([0,1)).
The correct answer is A. ([0,1)). The fractional part is never less than (0) and is always less than (1). So the range is ([0,1)).
भिन्नात्मक भाग हमेशा (0) से कम नहीं और (1) से कम होता है। इसलिए परिसर ([0,1)) है।
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What is the (y)-intercept of (f(x)=2x+3)?
B. (3)
Putting (x=0) gives (y=3). In (y=mx+c), (c) is the (y)-intercept.
The correct answer is B. (3). Putting (x=0) gives (y=3). In (y=mx+c), (c) is the (y)-intercept.
(x=0) रखने पर (y=3) मिलता है। (y=mx+c) में (c) (y)-प्रतिच्छेद है।
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What is the zero of (f(x)=3x-9)?
B. (x=3)
For the zero, set (3x-9=0). This gives (x=3).
The correct answer is B. (x=3). For the zero, set (3x-9=0). This gives (x=3).
शून्यक के लिए (3x-9=0) रखें। इससे (x=3) मिलता है।
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What is the slope of (f(x)=-2x+4)?
C. (-2)
In (y=mx+c), (m) is the slope. Here (m=-2).
The correct answer is C. (-2). In (y=mx+c), (m) is the slope. Here (m=-2).
(y=mx+c) में (m) ढाल होता है। यहां (m=-2) है।
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What are the (x)-intercepts of (f(x)=x-2-1)?
A. (x=1) और (x=-1)(x=1) and (x=-1)
From \(x^2-1=0\), we get \(x^2=1\). So \(x=\pm 1\) are the intercepts.
The correct answer is A. (x=1) और (x=-1) / (x=1) and (x=-1). From \(x^2-1=0\), we get \(x^2=1\). So \(x=\pm 1\) are the intercepts.
\(x^2-1=0\) से \(x^2=1\) मिलता है। इसलिए \(x=\pm 1\) प्रतिच्छेद हैं।
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What is the vertex of (f(x)=x-2+2x+1)?
A. ((-1,0))
(x-2+2x+1=(x+1)2). Therefore the vertex is ((-1,0)).
The correct answer is A. ((-1,0)). (x-2+2x+1=(x+1)2). Therefore the vertex is ((-1,0)).
(x-2+2x+1=(x+1)2)। इसलिए शीर्ष ((-1,0)) है।
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What is the minimum value of (f(x)=x-2-2x+1)?
A. (0)
(x-2-2x+1=(x-1)2), so the minimum value is (0). It occurs at (x=1).
The correct answer is A. (0). (x-2-2x+1=(x-1)2), so the minimum value is (0). It occurs at (x=1).
(x-2-2x+1=(x-1)2), इसलिए न्यूनतम मान (0) है। यह (x=1) पर मिलता है।
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What is the vertex of (f(x)=|x-2|+1)?
A. ((2,1))
(|x-2|) becomes zero at (x=2) and (1) is added outside. So the vertex is ((2,1)).
The correct answer is A. ((2,1)). (|x-2|) becomes zero at (x=2) and (1) is added outside. So the vertex is ((2,1)).
(|x-2|) शून्य (x=2) पर होता है और बाहर (1) जुड़ता है। इसलिए शीर्ष ((2,1)) है।
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What is the minimum value of (f(x)=|x+3|-2)?
C. (-2)
The minimum value of (|x+3|) is (0). Therefore the minimum value of (|x+3|-2) is (-2).
The correct answer is C. (-2). The minimum value of (|x+3|) is (0). Therefore the minimum value of (|x+3|-2) is (-2).
(|x+3|) का न्यूनतम मान (0) है। इसलिए (|x+3|-2) का न्यूनतम मान (-2) है।
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What is the maximum value of (f(x)=-|x|+6)?
B. (6)
Since \(-|x|\le 0\), \(-|x|+6\le 6\). The maximum value (6) occurs at (x=0).
The correct answer is B. (6). Since \(-|x|\le 0\), \(-|x|+6\le 6\). The maximum value (6) occurs at (x=0).
क्योंकि \(-|x|\le 0\), इसलिए \(-|x|+6\le 6\)। अधिकतम मान (6) (x=0) पर मिलता है।
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What is the domain of (f(x)=x-3)?
A. सभी वास्तविक (x)All real (x)
\(x^3\) is defined for every real (x). Therefore its domain is all real numbers.
The correct answer is A. सभी वास्तविक (x) / All real (x). \(x^3\) is defined for every real (x). Therefore its domain is all real numbers.
\(x^3\) हर वास्तविक (x) के लिए परिभाषित है। इसलिए इसका डोमेन सभी वास्तविक संख्याएं हैं।
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What is the (y)-intercept of (f(x)=x-3)?
A. (0)
Putting (x=0), (f(0)=03=0). Therefore the (y)-intercept is (0).
The correct answer is A. (0). Putting (x=0), (f(0)=03=0). Therefore the (y)-intercept is (0).
(x=0) रखने पर (f(0)=03=0)। इसलिए (y)-अक्ष प्रतिच्छेद (0) है।
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What is the (x)-intercept of (f(x)=x-3+8)?
B. (x=-2)
On the (x)-axis, \(x^3+8=0\), so \(x^3=-8\). This gives (x=-2).
The correct answer is B. (x=-2). On the (x)-axis, \(x^3+8=0\), so \(x^3=-8\). This gives (x=-2).
(x)-अक्ष पर \(x^3+8=0\), इसलिए \(x^3=-8\)। इससे (x=-2) मिलता है।
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What is the starting point of (f(x)=\sqrt{x-2}+5)?
A. ((2,5))
\(\sqrt{x-2}\) starts at (x=2), and (5) is added outside. So the starting point is ((2,5)).
The correct answer is A. ((2,5)). \(\sqrt{x-2}\) starts at (x=2), and (5) is added outside. So the starting point is ((2,5)).
\(\sqrt{x-2}\) (x=2) से शुरू होता है और बाहर (5) जुड़ता है। इसलिए प्रारंभिक बिंदु ((2,5)) है।
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What is the starting point of (f(x)=\sqrt{x+1}-3)?
A. ((-1,-3))
\(\sqrt{x+1}\) starts at (x=-1), and (3) is subtracted outside. So the starting point is ((-1,-3)).
The correct answer is A. ((-1,-3)). \(\sqrt{x+1}\) starts at (x=-1), and (3) is subtracted outside. So the starting point is ((-1,-3)).
\(\sqrt{x+1}\) (x=-1) से शुरू होता है और बाहर (3) घटता है। इसलिए प्रारंभिक बिंदु ((-1,-3)) है।
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What is the vertical asymptote of (f(x)=\frac{1}{x+5})?
B. (x=-5)
The function is not defined when the denominator (x+5) is zero. Therefore (x=-5) is the vertical asymptote.
The correct answer is B. (x=-5). The function is not defined when the denominator (x+5) is zero. Therefore (x=-5) is the vertical asymptote.
हर (x+5) शून्य होने पर फलन परिभाषित नहीं होता। इसलिए (x=-5) ऊर्ध्व आसिम्प्टोट है।
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What is the horizontal asymptote of (f(x)=\frac{1}{x-1}-2)?
B. (y=-2)
For large (|x|), \(\frac{1}{x-1}\) approaches (0). So the horizontal asymptote is (y=-2).
The correct answer is B. (y=-2). For large (|x|), \(\frac{1}{x-1}\) approaches (0). So the horizontal asymptote is (y=-2).
बड़े (|x|) पर \(\frac{1}{x-1}\) (0) के पास जाता है। इसलिए क्षैतिज आसिम्प्टोट (y=-2) है।
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Which function has a graph symmetric about the (y)-axis?
A. (f(x)=x-2+1)
(f(-x)=(-x)2+1=x-2+1=f(x)). Therefore it is symmetric about the (y)-axis.
The correct answer is A. (f(x)=x-2+1). (f(-x)=(-x)2+1=x-2+1=f(x)). Therefore it is symmetric about the (y)-axis.
(f(-x)=(-x)2+1=x-2+1=f(x))। इसलिए यह (y)-अक्ष के प्रति सममित है।
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Which function has a graph symmetric about the origin?
A. (f(x)=x-3)
(f(-x)=(-x)3=-x-3=-f(x)). So the graph of \(x^3\) is symmetric about the origin.
The correct answer is A. (f(x)=x-3). (f(-x)=(-x)3=-x-3=-f(x)). So the graph of \(x^3\) is symmetric about the origin.
(f(-x)=(-x)3=-x-3=-f(x))। इसलिए \(x^3\) का आलेख मूलबिंदु के प्रति सममित है।
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Which graph will fail the vertical line test?
C. \(x^2+y^2=4\)
The circle \(x^2+y^2=4\) gives two (y)-values for many (x)-values. So it is not the graph of a function.
The correct answer is C. \(x^2+y^2=4\). The circle \(x^2+y^2=4\) gives two (y)-values for many (x)-values. So it is not the graph of a function.
वृत्त \(x^2+y^2=4\) में कई (x)-मानों पर दो (y)-मान मिलते हैं। इसलिए यह फलन का आलेख नहीं है।
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What is the domain of (f(x)=\sqrt{9-x})?
A. \(x\le 9\)
For a real square root, \(9-x\ge 0\) is required, so \(x\le 9\). In square root graphs, keep the inside expression (0) or positive.
The correct answer is A. \(x\le 9\). For a real square root, \(9-x\ge 0\) is required, so \(x\le 9\). In square root graphs, keep the inside expression (0) or positive.
वास्तविक वर्गमूल के लिए \(9-x\ge 0\) चाहिए, इसलिए \(x\le 9\)। वर्गमूल वाले ग्राफ में अंदर की राशि को (0) या धनात्मक रखें।
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