Question

sketch the graph of f(x) = 2x²/x²+2x-3
and give its domain intercepts asymptotes​

Answer 1

Answer:

How do you sketch the graph of f(x) = (2x^2) / (X^2+2x-3) and give its domain, intercepts, and asymptotes?

f(x)=2x2x2+2x−3=2x2(x−1)(x+3).

The denominator becomes zero when x=1 or x=−3.

⇒ The domain of the function is R−−3,1.

The numerator is always non negative. Hence the value of the function take the same sign as that of the denominator.

⇒ The function is positive in interval (−∞,−3)∪(1,∞) and negative in the interval (−3,1).

⇒ The function tends to ∞ as x tends to −3− and 1+ and the function tends to −∞ as x tends to −3+ and 1−.

⇒ The function has vertical asymptotes at x=1 and x=−3.

f(x)=2x2x2+2x−3=21+2/x−3/x2.

⇒ The function tends to 2 as x tends to ±∞ .

⇒ The function has horizontal asymptote at y=2 .

The value of the function becomes zero at x=0 and at no other point.

⇒ The function has a root at x=0.

f′(x)=4x(x−3)(x−1)2(x+3)2.

When f′(x)=0,x=0 or x=3.

At x=0,f(x)=0 and at x=3,f(x)=1.5.

f(0−)<0 and f(0+)<0.

⇒f(x) has a local maximum at x=0.

f(3−)>1.5 and f(3+)>1.5.

⇒f(x) has a local minimum at x=3.

The function touches the origin while passing from the third quadrant to the fourth quadrant and does not have any other intercept with either the X axis or the Y axis.

Answer 2

Given:

The sketch of the graph graph of $f(x) = \frac{2x}^{2}{x}^{2}+2x-3$.

To find:

The domain intercepts asymptotes​.

Step-by-step explanation:

$f(x) =\frac {2x}^{2}{x}^{2}+2x-3$

$=\frac{2x^{2} }{(x-1)(x+3)}$

The denominator becomes zero when $x=1orx=-3$

$f(x) = \frac{2x}^{2}{x}^{2}+2x-3$

$=\frac{2}{1+2/x-3/x^{2}}$

The Function has horizontal asymptotes​ $y=2$.

The function has a root at $x=0$.

$f^{'}(x)=\frac{4x(x-3)}{(x-1)^{2}(x+3)^{2}}$

$f'(x)=0,X=0orX=3$

$x=3,f(x)=1.5$

$f(x)$ has a local maximum at $x=0$

$f(x)$has a local minimum at $x=3$

Answer:

Therefore, The domain intercepts asymptotes $x=0,x=3$.​