Mathematics, 22.02.2020 23:52 lee1677

A portion of the graph of f(x) = −x2 + 6x − 5 is shown. Which of the following describes all solutions for f(x)? a partial parabola opening down and passing through 1 comma 0 and 5 comma zero with endpoints at 0 comma negative 5 and 6 comma negative 5 (x, y) for all real numbers (x, −x2 + 6x − 5) where 0 ≤ x ≤ 6 (x, −x2 + 6x − 5) where 1 ≤ x ≤ 5 (x, y) where 1 ≤ x ≤ 5

Answers: 2

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Answers

Answer from: tlturner03

Step-by-step explanation:

Considering the function

$f\left(x\right)=-x^2+6x-5$

solving

-x^2+6x-5=0

$\mathrm{Quadratic\:Equation\:Formula:}$

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=-1,\:b=6,\:c=-5:\quad x_{1,\:2}=\frac{-6\pm \sqrt{6^2-4\left(-1\right)\left(-5\right)}}{2\left(-1\right)}$

$x=\frac{-6+\sqrt{6^2-4\left(-1\right)\left(-5\right)}}{2\left(-1\right)}$

$=\frac{-6+\sqrt{6^2-4\cdot \:1\cdot \:5}}{-2\cdot \:1}$

$=\frac{-6+\sqrt{16}}{-2\cdot \:1}$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{-b}=\frac{a}{b}$

$-6+\sqrt{16}=-\left(6-\sqrt{16}\right)$

$=\frac{6-\sqrt{16}}{2}$

$=\frac{6-4}{2}$

x=1

Similarly,

$x=\frac{-6-\sqrt{6^2-4\left(-1\right)\left(-5\right)}}{2\left(-1\right)}$

$=\frac{-6-\sqrt{6^2-4\cdot \:1\cdot \:5}}{-2\cdot \:1}$

$=\frac{-6-\sqrt{16}}{-2}$

$=\frac{6+\sqrt{16}}{2}$

$\:x=5$

$\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}$

$x=1,\:x=5$

Also check the attached graph below.

From the graph as shown below, it is clear that

$\mathrm{Domain\:of\:}\:-x^2+6x-5\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:$

and

$\mathrm{Range\:of\:}-x^2+6x-5:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\le \:4\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:4]\end{bmatrix}$

Also

$\mathrm{Axis\:interception\:points\:of}\:-x^2+6x-5:\quad \mathrm{X\:Intercepts}:\:\left(1,\:0\right),\:\left(5,\:0\right),\:\mathrm{Y\:Intercepts}:\:\left(0,\:-5\right)$

$\mathrm{Vertex\:of}\:-x^2+6x-5:\quad \mathrm{Maximum}\space\left(3,\:4\right)$

Thus, from the figure, it is clear that a partial parabola opening down, passing through (1, 0) and (5, 0) with endpoints (0, -5) and (6, -5).

A portion of the graph of f(x) = −x2 + 6x − 5 is shown. Which of the following describes all solutio

Answer from: Quest

if x = 1, then the answer is 3.5

Answer from: Quest

part a

answer: a reasonable domain is $0 \le n \le 10$ where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

=====================================

part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

=====================================

part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).