3.9 Modeling using variation  (Page 7/14)

$f(x)=x^5+4x^4+4x^3$

$f(x)=x^3-4x^2+x-4$

Use the Intermediate Value Theorem to show that at least one zero lies between 2 and 3 for the function $f(x)=x^3-5x+1$

$\frac{x^3-2x^2+4x+4}{x-2}$

$\frac{3x^4-4x^2+4x+8}{x+1}$

$\frac{x^3-2x^2+5x-1}{x+3}$

$\frac{x^3+4x+10}{x-3}$

$\frac{2x^3+6x^2-11x-12}{x+4}$

$\frac{3x^4+3x^3+2x+2}{x+1}$

$2x^3-3x^2-18x-8=0$

$3x^3+11x^2+8x-4=0$

$2x^4-17x^3+46x^2-43x+12=0$

$4x^4+8x^3+19x^2+32x+12=0$

$x^3-3x^2-2x+4=0$

$2x^4-x^3+4x^2-5x+1=0$

$f(x)=\frac{x+2}{x-5}$

$f(x)=\frac{x^2+1}{x^2-4}$

$f(x)=\frac{3x^2-27}{x^2+x-2}$

$f(x)=\frac{x+2}{x^2-9}$

$f(x)=\frac{x^2-1}{x+2}$

$f(x)=\frac{2x^3-x^2+4}{x^2+1}$

$f(x)={(x-2)}^2,x\ge 2$

$f(x)={(x+4)}^2-3,x\ge -4$

$f(x)=x^2+6x-2,x\ge -3$

$f(x)=2x^3-3$

$f(x)=\sqrt{4x+5}-3$

$f(x)=\frac{x-3}{2x+1}$

$y$ varies directly as the square of $x.$ If when $x=3,y=36,$ find $y$ if $x=4.$

$y$ varies inversely as the square root of $x$ If when $x=25,y=2,$ find $y$ if $x=4.$

$y$ varies jointly as the cube of $x$ and as $z.$ If when $x=1$ and $z=2,$ $y=6,$ find $y$ if $x=2$ and $z=3.$

$y$ varies jointly as $x$ and the square of $z$ and inversely as the cube of $w.$ If when $x=3,$ $z=4,$ and $w=2,$ $y=48,$ find $y$ if $x=4,$ $z=5,$ and $w=3.$

The weight of an object above the surface of the earth varies inversely with the distance from the center of the earth. If a person weighs 150 pounds when he is on the surface of the earth (3,960 miles from center), find the weight of the person if he is 20 miles above the surface.

The volume $V$ of an ideal gas varies directly with the temperature $T$ and inversely with the pressure P. A cylinder contains oxygen at a temperature of 310 degrees K and a pressure of 18 atmospheres in a volume of 120 liters. Find the pressure if the volume is decreased to 100 liters and the temperature is increased to 320 degrees K.

$\left(3-4i\right)\left(4+2i\right)$

$\frac{1-4i}{3+4i}$

$x^2-4x+13=0$

$f(x)=x^3\left(3-6x^2-2x^2\right)$

$f(x)=8x^3-3x^2+2x-4$

$f(x)=-2x^2(4-3x-5x^2)$

$f(x)=x^2+2x-8$

Vertex $(2,0)$ and point on graph $(4,12).$

A rectangular field is to be enclosed by fencing. In addition to the enclosing fence, another fence is to divide the field into two parts, running parallel to two sides. If 1,200 feet of fencing is available, find the maximum area that can be enclosed.

$f(x)={(x-3)}^3(3x-1){(x-1)}^2$

$f(x)=2x^6-6x^5+18x^4$

$\frac{2x^3+3x-4}{x+2}$

$\frac{x^4+3x^2-4}{x-2}$

$\frac{2x^3+5x^2-7x-12}{x+3}$

$f(x)=2x^3+5x^2-6x-9$

$f(x)=4x^4+8x^3+21x^2+17x+4$

$f(x)=4x^4+16x^3+13x^2-15x-18$

$f(x)=x^5+6x^4+13x^3+14x^2+12x+8$

It has a double zero at $x=3$ and zeroes at $x=1$ and $x=-2$ . It’s y -intercept is $(0,12).$

It has a zero of multiplicity 3 at $x=\frac{1}{2}$ and another zero at $x=-3$ . It contains the point $(1,8).$

$8x^3-21x^2+6=0$

$f(x)=\frac{x+4}{x^2-2x-3}$

$f(x)=\frac{x^2+2x-3}{x^2-4}$

$f(x)=\frac{x^2+3x-3}{x-1}$

$f(x)=\sqrt{x-2}+4$

$f(x)=3x^3-4$

$f(x)=\frac{2x+3}{3x-1}$

$y$ varies inversely as the square of $x$ and when $x=3,$ $y=2.$ Find $y$ if $x=1.$

$y$ varies jointly with $x$ and the cube root of $z.$ If when $x=2$ and $z=27,$ $y=12,$ find $y$ if $x=5$ and $z=8.$

The distance a body falls varies directly as the square of the time it falls. If an object falls 64 feet in 2 seconds, how long will it take to fall 256 feet?