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name kuta software - infinite algebra 2 date_______ period____ graphing…

Question

name
kuta software - infinite algebra 2
date_____ period__
graphing polynomial functions
state the maximum number of turns the graph of each function could make. then sketch the graph. state
the number of real zeros. approximate each zero to the nearest tenth. approximate the relative minima
and relative maxima to the nearest tenth.

  1. $f(x)=x^{2}+2x-5$
  2. $f(x)=-x^{4}+x^{3}+2x^{2}$
  3. $f(x)=x^{4}-4x^{3}+2x^{2}+x+4$
  4. $f(x)=x^{3}+x^{2}-x-2$

Explanation:

Response
Problem 1: $f(x)=x^2 + 2x - 5$

Step1: Find max turns (degree-1)

Max turns $= 2-1=1$

Step2: Find real zeros (quadratic formula)

$x=\frac{-2\pm\sqrt{2^2-4(1)(-5)}}{2(1)}=\frac{-2\pm\sqrt{24}}{2}=-1\pm\sqrt{6}$
Approximate: $x\approx1.4$, $x\approx-3.4$

Step3: Find relative extrema (vertex)

Vertex $x=-\frac{b}{2a}=-\frac{2}{2(1)}=-1$
$f(-1)=(-1)^2+2(-1)-5=-6$
Relative min at $(-1, -6)$

Step1: Find max turns (degree-1)

Max turns $=4-1=3$

Step2: Factor to find zeros

$f(x)=-x^2(x^2 - x - 2)=-x^2(x-2)(x+1)$
Zeros: $x=0$ (double root), $x=2$, $x=-1$

Step3: Find relative extrema (derivative)

$f'(x)=-4x^3+3x^2+4x$
Set $f'(x)=0$: $-x(4x^2-3x-4)=0$
Solve $4x^2-3x-4=0$: $x=\frac{3\pm\sqrt{9+64}}{8}=\frac{3\pm\sqrt{73}}{8}$
Approximate: $x\approx1.4$, $x\approx-0.7$
Evaluate $f(x)$:
$f(1.4)\approx-(1.4)^4+(1.4)^3+2(1.4)^2\approx3.0$
$f(-0.7)\approx-(-0.7)^4+(-0.7)^3+2(-0.7)^2\approx0.4$
$f(0)=0$

Step1: Find max turns (degree-1)

Max turns $=4-1=3$

Step2: Approximate real zeros (graph/root test)

Test values: $f(1)=1-4+2+1+4=4$, $f(2)=16-32+8+2+4=-2$, $f(3)=81-108+18+3+4=-2$, $f(4)=256-256+32+4+4=40$
Use linear approximation:
Between $x=1$ and $x=2$: $x\approx1.8$
Between $x=3$ and $x=4$: $x\approx3.2$

Step3: Find relative extrema (derivative)

$f'(x)=4x^3-12x^2+4x+1$
Approximate roots of $f'(x)=0$: $x\approx0.3$, $x\approx0.9$, $x\approx2.8$
Evaluate $f(x)$:
$f(0.3)\approx(0.3)^4-4(0.3)^3+2(0.3)^2+(0.3)+4\approx4.2$
$f(0.9)\approx(0.9)^4-4(0.9)^3+2(0.9)^2+(0.9)+4\approx4.0$
$f(2.8)\approx(2.8)^4-4(2.8)^3+2(2.8)^2+(2.8)+4\approx-2.1$

Answer:

Maximum number of turns: 1
Real zeros: $x\approx1.4$, $x\approx-3.4$
Relative minimum: $(-1, -6)$
(Graph: Opens upward parabola with vertex at $(-1,-6)$, crossing x-axis at the approximate zeros)

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