Question

I was wondering if you could help me with my homework

Answer 1

f(x) is undefined for all x < 0.

Hence, the domain of f is [0, ∞)

`f(0)=\sqrt[]{0^2+6(0)}-(0)=0`

Hence, the y-intercept is 0

When f(x) = 0

`\begin{gathered} \sqrt[]{x^2+6x}-x=0 \\ \sqrt[]{x^2+6x}=x \\ \text{ Squaring both sides, we have} \\ x^2+6x=x^2 \\ \text{Hence} \\ 6x=0 \\ (6x)/(6)=(0)/(6) \\ \text{Thus x = 0} \end{gathered}`

Hence, the x-intercept is also 0

The image of the graph is as shown below

From the image of the graph, we can see that there is no apparent symmetry

`\begin{gathered} f(x)=\sqrt[]{x^2+6x}-x \\ \text{ Therefore,} \\ f^(\prime)(x)=(1)/(2)*(2x+6)(x^2+6x)^{-(1)/(2)}-1 \end{gathered}`
`f^(\prime)(x)=((x+3))/(x(x+6))-1`

Therefore.

`y^(\prime)(*)=\frac{(x+3)}{√(x(x+6))^{}}-1`
`y^{^{^(\doubleprime)}}(x)=\text{ }-(9)/(\left(x^2+6x\right)√(x^2+6x))`

From the graph, we can see that there is no vertical asymptote,

But there is a horizontal asymptote, y = 3

The graph f(x) is increasing for f'(x) > 0

`\begin{gathered} ^(\prime)((x+3))/(x(x+6))-1>0 \\ x+3-x^2-6x>0 \\ x^2+5x-3<0 \\ \end{gathered}`

From the image, we can see that the function is increasing on the interval [0, ∞ )