Question

for questions 20 - 21, find the intercepts and relative extrema for the graph of each function. 21. (h(x)=2x^{3}+5x^{2}-25x) (1 point)

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 \(x=\frac{-5 + 5\sqrt{7}}{6}\)