Summarize the pert nent information obleined by applying the graphing stralegy and sketch the graph of $$ y=f(x) $$. $$ f(x)=\left(x^{2}+10\right)\left(100-x^{2}\right) $$ Select the correct choice below and, if nocessery, fill in the answer box to complete your choice. A. The $$ x $$-inlercapt(s) of $$ f $$ is/are $$ x=-10,10 $$. (Type an exact answer, using radicals as needed. Use a comma to separate answers as needed.) B. The function $$ f $$ has no k -intercepts. Select the correct choico below and, if necessary, fill in the answer box to complete your choice. A. The $$ y $$-inlercapt of $$ f $$ is $$ y=1000 $$. (Type an exact answer, using radicals as needed.) B. The function $$ f $$ has no $$ y $$ inlercept. Solect the correct choico bolow and, if necessery, fill in the answer box to complete your choice. A. The function fis increasing on the subinterval(s) $$ \square $$ (Type your answer in interval notation. Type an exacl answer, using racicals as needed. Use a comma lo separale answers as needed.) B. The function fis never increasing.

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The function is $$ f(x)=\left(x^2+10\right)\left(100-x^2\right) $$ which can be expanded as $$ f(x)=-x^4+90x^2+1000. $$ **Step 1. Finding the $$x$$-intercepts** Set $$ f(x)=0 $$: $$ (x^2+10)(100-x^2)=0. $$ The first factor gives $$ x^2+10=0 \quad \Longrightarrow \quad x^2=-10, $$ which has no real solutions. The second factor gives $$ 100-x^2=0 \quad \Longrightarrow \quad x^2=100, $$ so $$ x=\pm 10. $$ Thus, the correct choice is A. The $$x$$-intercepts of $$ f $$ is/are $$x=-10,10$$. **Step 2. Finding the $$y$$-intercept** Evaluate $$ f(0) $$: $$ f(0)=(0^2+10)(100-0^2)=10\cdot 100=1000. $$ Thus, the correct choice is A. The $$y$$-intercept of $$ f $$ is $$y=1000$$. **Step 3. Determining where the function is increasing** First, compute the derivative. It is easier to differentiate the expanded form: $$ f(x)=-x^4+90x^2+1000. $$ Thus, $$ f'(x)=-4x^3+180x. $$ Factor the derivative: $$ f'(x)=-4x(x^2-45)=-4x \left(x-\sqrt{45}\right)\left(x+\sqrt{45}\right). $$ Since $$\sqrt{45}=3\sqrt{5}$$, the critical points are $$ x=-3\sqrt{5},\quad x=0,\quad x=3\sqrt{5}. $$ Now, analyze the sign of $$ f'(x) $$: - For $$ x<-3\sqrt{5} $$: Choose a test value (e.g. $$x=-10$$). $$-4x>0$$ (since $$x$$ is negative) and $$(x^2-45)>0$$. Therefore, $$f'(x)>0$$. - For $$-3\sqrt{5}0$$ but $$(x^2-45)<0$$, so $$f'(x)<0$$. - For $$00$$. - For $$ x>3\sqrt{5} $$: Choose a test value (e.g. $$x=10$$). $$-4(10)<0$$ and $$(x^2-45)>0$$, hence $$f'(x)<0$$. Thus, $$ f $$ is increasing on $$ (-\infty,-3\sqrt{5}) \quad \text{and} \quad (0, 3\sqrt{5}). $$ So the correct choice is A. The function $$ f $$ is increasing on the subinterval(s) $$(-\infty,-3\sqrt{5})$$ and $$(0,3\sqrt{5})$$. **Final Answers:** 1. $$x$$-intercepts: $$x=-10, 10$$. 2. $$y$$-intercept: $$y=1000$$. 3. Increasing on: $$ (-\infty,-3\sqrt{5})\cup (0,3\sqrt{5}). $$ The function $$ f(x) = (x^2 + 10)(100 - x^2) $$ has: - $$ x $$-intercepts at $$ x = -10 $$ and $$ x = 10 $$. - A $$ y $$-intercept at $$ y = 1000 $$. - It is increasing on the intervals $$ (-\infty, -3\sqrt{5}) $$ and $$ (0, 3\sqrt{5}) $$.

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