x3+1
Q4. Trace the curve giving all the necessary details about the curve: y
x
Answers
Answer:
Trace the curve y=x
Trace the curve y=x 3
Trace the curve y=x 3 .
Trace the curve y=x 3 .1. Domain, extent, intercept and origin; when x∈R, y is well defined.
Trace the curve y=x 3 .1. Domain, extent, intercept and origin; when x∈R, y is well defined. As x→+∞; y→±∞, the curve exists in first and fourth quadrant only.
Trace the curve y=x 3 .1. Domain, extent, intercept and origin; when x∈R, y is well defined. As x→+∞; y→±∞, the curve exists in first and fourth quadrant only. The intercepts with the axes are given by x=0, y=0 and when y=0, x=0
Trace the curve y=x 3 .1. Domain, extent, intercept and origin; when x∈R, y is well defined. As x→+∞; y→±∞, the curve exists in first and fourth quadrant only. The intercepts with the axes are given by x=0, y=0 and when y=0, x=0 ⇒ the curve passes through the origin.
Trace the curve y=x 3 .1. Domain, extent, intercept and origin; when x∈R, y is well defined. As x→+∞; y→±∞, the curve exists in first and fourth quadrant only. The intercepts with the axes are given by x=0, y=0 and when y=0, x=0 ⇒ the curve passes through the origin.2. Symmetry : By symmetry test, we have the curve is symmetric about origin.
Trace the curve y=x 3 .1. Domain, extent, intercept and origin; when x∈R, y is well defined. As x→+∞; y→±∞, the curve exists in first and fourth quadrant only. The intercepts with the axes are given by x=0, y=0 and when y=0, x=0 ⇒ the curve passes through the origin.2. Symmetry : By symmetry test, we have the curve is symmetric about origin.3. Asymptotes : As x→+∞, y→+∞ and vice versa.
Trace the curve y=x 3 .1. Domain, extent, intercept and origin; when x∈R, y is well defined. As x→+∞; y→±∞, the curve exists in first and fourth quadrant only. The intercepts with the axes are given by x=0, y=0 and when y=0, x=0 ⇒ the curve passes through the origin.2. Symmetry : By symmetry test, we have the curve is symmetric about origin.3. Asymptotes : As x→+∞, y→+∞ and vice versa. ∴ the curve does not admit asymptotes.
Trace the curve y=x 3 .1. Domain, extent, intercept and origin; when x∈R, y is well defined. As x→+∞; y→±∞, the curve exists in first and fourth quadrant only. The intercepts with the axes are given by x=0, y=0 and when y=0, x=0 ⇒ the curve passes through the origin.2. Symmetry : By symmetry test, we have the curve is symmetric about origin.3. Asymptotes : As x→+∞, y→+∞ and vice versa. ∴ the curve does not admit asymptotes.4. Monotonicity :
Trace the curve y=x 3 .1. Domain, extent, intercept and origin; when x∈R, y is well defined. As x→+∞; y→±∞, the curve exists in first and fourth quadrant only. The intercepts with the axes are given by x=0, y=0 and when y=0, x=0 ⇒ the curve passes through the origin.2. Symmetry : By symmetry test, we have the curve is symmetric about origin.3. Asymptotes : As x→+∞, y→+∞ and vice versa. ∴ the curve does not admit asymptotes.4. Monotonicity : dx
Trace the curve y=x 3 .1. Domain, extent, intercept and origin; when x∈R, y is well defined. As x→+∞; y→±∞, the curve exists in first and fourth quadrant only. The intercepts with the axes are given by x=0, y=0 and when y=0, x=0 ⇒ the curve passes through the origin.2. Symmetry : By symmetry test, we have the curve is symmetric about origin.3. Asymptotes : As x→+∞, y→+∞ and vice versa. ∴ the curve does not admit asymptotes.4. Monotonicity : dxdy
Trace the curve y=x 3 .1. Domain, extent, intercept and origin; when x∈R, y is well defined. As x→+∞; y→±∞, the curve exists in first and fourth quadrant only. The intercepts with the axes are given by x=0, y=0 and when y=0, x=0 ⇒ the curve passes through the origin.2. Symmetry : By symmetry test, we have the curve is symmetric about origin.3. Asymptotes : As x→+∞, y→+∞ and vice versa. ∴ the curve does not admit asymptotes.4. Monotonicity : dxdy
=3x 2 , hence ∀x, dxdy
=3x 2 , hence ∀x, dxdy >0
=3x 2 , hence ∀x, dxdy >0 ∴ the graph is monotonically increasing and will not intersect the axes at points other than origin.
=3x 2 , hence ∀x, dxdy >0 ∴ the graph is monotonically increasing and will not intersect the axes at points other than origin.5. dx d 2 y =6x is 0 for x=0. Special point (0,0) is a point of inflection.