Question

ap precalculus
name: elfidio mejía 9/22/25
quarter 1 week 5. this warm - up is due friday september 26 during the first five minutes of class. answer each problem completely showing all work in the space provided and placing your answer in the space provided. if you do not show any work for a problem, then you will not receive any credit for that problem. this assignment will be recorded in the classwork category. if you are absent in friday, this is due the first day you are back in class. this assignment is not eligible for retake.
let f(x)=368x² - 15x³ - 2x⁴+1725x - 6300 (may use calculator for #2 and #3)

1. determine if f(x) has an absolute minimum or maximum or neither. justify your answer.
2. find all the real zeros and the y - intercept of f(x).
3. on the closed interval -15,15, find the absolute minimum output value of f(x).
4. let k(x)=35 + 11x - 6x², find all values of x for which k(x)≤0. (no calculator)
5. find a linear equation d(t) in slope intercept form such that d(-2)=5 and d(1)=3. use simplified improper fractions

Explanation:

Step1: Analyze the degree and leading - coefficient of \(f(x)\)

The function \(f(x)=- 2x^{4}-15x^{3}+368x^{2}+1725x - 6300\) is a polynomial of degree \(n = 4\) (even) and leading - coefficient \(a=-2\lt0\). As \(x\to\pm\infty\), \(y = f(x)\to-\infty\). So, \(f(x)\) has no absolute minimum and has an absolute maximum.

Step2: Find the \(y\) - intercept of \(f(x)\)

Set \(x = 0\) in \(f(x)\). Then \(f(0)=-6300\). To find the real zeros, we can use the Rational Root Theorem. The possible rational roots are factors of \(\frac{6300}{2}=3150\). The factors of \(3150=2\times3^{2}\times5^{2}\times7\). The possible rational roots are \(\pm1,\pm2,\pm3,\pm4,\pm5,\pm6,\pm7,\pm9,\pm10,\pm12,\pm14,\pm15,\pm18,\pm20,\pm21,\pm25,\pm28,\pm30,\pm35,\pm42,\pm45,\pm50,\pm63,\pm70,\pm75,\pm84,\pm90,\pm105,\pm126,\pm140,\pm150,\pm175,\pm180,\pm210,\pm225,\pm252,\pm315,\pm350,\pm420,\pm450,\pm525,\pm630,\pm750,\pm1050,\pm1575,\pm3150\). Using a calculator or synthetic division, we can find the real roots.

Step3: Find the absolute minimum of \(f(x)\) on \([-15,15]\)

First, find the derivative \(f^\prime(x)=-8x^{3}-45x^{2}+736x + 1725\). Then find the critical points by setting \(f^\prime(x)=0\). Using a calculator, we can find the critical points in the interval \([-15,15]\). Evaluate \(f(x)\) at the critical points and at the endpoints \(x=-15\) and \(x = 15\).

• \(f(-15)=-2(-15)^{4}-15(-15)^{3}+368(-15)^{2}+1725(-15)-6300=-2\times50625+15\times3375 + 368\times225-25875-6300=-101250 + 50625+82800-25875 - 6300=0\)
• \(f(15)=-2(15)^{4}-15(15)^{3}+368(15)^{2}+1725(15)-6300=-2\times50625-15\times3375+368\times225 + 25875-6300=-101250-50625 + 82800+25875-6300=-4950\)

Step4: Solve \(k(x)=35 + 11x-6x^{2}\leq0\)

Rewrite \(k(x)\) as \(6x^{2}-11x - 35\geq0\). Factor \(6x^{2}-11x - 35=(2x - 7)(3x+5)\). Set \((2x - 7)(3x + 5)=0\), the roots are \(x=\frac{7}{2}\) and \(x=-\frac{5}{3}\). The solution of \(6x^{2}-11x - 35\geq0\) is \(x\leq-\frac{5}{3}\) or \(x\geq\frac{7}{2}\).

Step5: Find the linear equation \(d(t)=mt + b\)

The slope \(m=\frac{d(1)-d(-2)}{1-(-2)}=\frac{3 - 5}{3}=-\frac{2}{3}\).
Using the point - slope form with the point \((t_1,d(t_1))=(1,3)\), \(d(t)-3=-\frac{2}{3}(t - 1)\). Then \(d(t)=-\frac{2}{3}t+\frac{2}{3}+3=-\frac{2}{3}t+\frac{11}{3}\).

Answer:

1. \(f(x)\) has an absolute maximum and no absolute minimum because it is a degree - 4 polynomial with a negative leading coefficient.
2. \(y\) - intercept: \(-6300\). Real zeros need further calculation using the Rational Root Theorem and synthetic division or a calculator.
3. The absolute minimum value of \(f(x)\) on \([-15,15]\) is \(-4950\).
4. \(x\leq-\frac{5}{3}\) or \(x\geq\frac{7}{2}\).
5. \(d(t)=-\frac{2}{3}t+\frac{11}{3}\)