for questions 20 - 21, find the intercepts and relative extrema for the graph of each function. 21…

Explanation:

Step1: Find the x - intercepts

Set (h(x)=0), so (2x^{3}+5x^{2}-25x = 0). Factor out (x): (x(2x^{2}+5x - 25)=0). Then factor the quadratic: (x(2x - 5)(x+5)=0). Solving (x = 0), (2x-5=0) (gives (x=\frac{5}{2})), and (x + 5=0) (gives (x=-5)).

Step2: Find the y - intercept

Set (x = 0) in (h(x)). Then (h(0)=2(0)^{3}+5(0)^{2}-25(0)=0).

Step3: Find the derivative of (h(x))

Using the power - rule, (h^\prime(x)=6x^{2}+10x - 25).

Step4: Find the critical points

Set (h^\prime(x)=0). Using the quadratic formula (x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}) for (ax^{2}+bx + c = 0), here (a = 6), (b = 10), (c=-25). (x=\frac{-10\pm\sqrt{10^{2}-4\times6\times(-25)}}{2\times6}=\frac{-10\pm\sqrt{100 + 600}}{12}=\frac{-10\pm\sqrt{700}}{12}=\frac{-10\pm10\sqrt{7}}{12}=\frac{-5\pm5\sqrt{7}}{6}).

Step5: Determine relative extrema

We can use the second - derivative test. First, find (h^{\prime\prime}(x)=12x + 10). For (x=\frac{-5 + 5\sqrt{7}}{6}), (h^{\prime\prime}(\frac{-5 + 5\sqrt{7}}{6})=12(\frac{-5 + 5\sqrt{7}}{6})+10=-10 + 10\sqrt{7}+10=10\sqrt{7}>0), so there is a relative minimum at (x=\frac{-5 + 5\sqrt{7}}{6}). For (x=\frac{-5 - 5\sqrt{7}}{6}), (h^{\prime\prime}(\frac{-5 - 5\sqrt{7}}{6})=12(\frac{-5 - 5\sqrt{7}}{6})+10=-10-10\sqrt{7}+10=-10\sqrt{7}<0), so there is a relative maximum at (x=\frac{-5 - 5\sqrt{7}}{6}).

Answer:

• x - intercepts: (x = 0), (x=\frac{5}{2}), (x=-5)
• y - intercept: (y = 0)
• Relative maximum at (x=\frac{-5 - 5\sqrt{7}}{6})
• Relative minimum at (x=\frac{-5 + 5\sqrt{7}}{6})