f(x) is undefined for all x < 0.
Hence, the domain of f is [0, ∞)
[tex]f(0)=\sqrt[]{0^2+6(0)}-(0)=0[/tex]
Hence, the y-intercept is 0
When f(x) = 0
[tex]\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 \\ \frac{6x}{6}=\frac{0}{6} \\ \text{Thus x = 0} \end{gathered}[/tex]
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
[tex]\begin{gathered} f(x)=\sqrt[]{x^2+6x}-x \\ \text{ Therefore,} \\ f^{\prime}(x)=\frac{1}{2}\times(2x+6)(x^2+6x)^{-\frac{1}{2}}-1 \end{gathered}[/tex][tex]f^{\prime}(x)=\frac{(x+3)}{x(x+6)}-1[/tex]
Therefore.
[tex]y^{\prime}(\times)=\frac{(x+3)}{\sqrt{x(x+6)}^{}}-1[/tex][tex]y^{^{^{\doubleprime}}}(x)=\text{ }-\frac{9}{\left(x^2+6x\right)\sqrt{x^2+6x}}[/tex]
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
[tex]\begin{gathered} ^{\prime}\frac{(x+3)}{x(x+6)}-1>0 \\ x+3-x^2-6x>0 \\ x^2+5x-3<0 \\ \end{gathered}[/tex]
From the image, we can see that the function is increasing on the interval [0, ∞ )
