Question

Analyze the following quadratics. Be sure to include domain, range, extrema, zeros, y intercepts, increasing, decreasing, end behavior, and continuity.

Answer 1

Good luck on your homework.

Answer 2

`\begin{gathered} \text{domain: x }\epsilon R_{} \\ range\colon y\ge-(9)/(4) \\ \text{extrrema: 18} \\ \text{zeros} \\ x=3,\text{ x=6} \\ y-intercept\colon18 \\ \text{decreasing}(-\infty,4.5) \\ \text{ increasing:(4.5,}\infty) \\ \text{ inflection at x=4.5} \\ \text{continuos for all x} \end{gathered}`

Step-by-step explanation

`f(x)=x^2-9x+18`

Step 1

a)Domain:

The domain of a function is the set of all possible inputs for the function, in other words, the values that x can take, so

there is no restriction for this function, hence, the domain is all real numbers

`\text{Domain:(-}\infty,\infty)`

b)Range: is the set of all possible values that the function will give when we give in the domain as input

to know the range of a quadratic equation, do

a) set the function equals zero

`\begin{gathered} y=x^2-9x+18 \\ x^2-9x+18-y=0 \end{gathered}`

now, we need to solve for x, use the quadratic formula

let

`\begin{gathered} x^2-9x+18-y=0\rightarrow ax^2+bx+c \\ so \\ a=1 \\ b=-9 \\ c=18-y \end{gathered}`

apply the formula

`\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ x=\frac{-(-9)\pm\sqrt[]{(-9)^2-4(1)(18-y)}}{2\cdot1} \\ x=\frac{+9\pm\sqrt[]{81-72+4y}}{2} \\ x=\frac{+9\pm\sqrt[]{9+4y}}{2} \end{gathered}`

now, this has solution only if the value insides the roor sign, is greater or equal than zero,so

`\begin{gathered} 9+4y\ge0 \\ \text{subtract 9 in both sides} \\ 9+4y-9\ge0-9 \\ 4y\ge-9 \\ \text{divide both sides by 4} \\ (4y)/(4)\ge-(9)/(4) \\ y\ge-(9)/(4) \end{gathered}`

hence, the range is

`\begin{gathered} y\ge-(9)/(4) \\ \lbrack-(9)/(4),\infty) \end{gathered}`

Step 2

extremes A quadratic function f(x)=ax2+bx+c has an extreme value at its vertex, so if a>0 , then f(−ba) is the maximum, and if a<0

so

`\begin{gathered} f(x)=x^2-9x+18 \\ a=1 \\ -ba=-(-9)\cdot1=9 \\ f(-ba)=9^2-9\cdot9+18 \\ f(-ba)=81-81+18 \\ f(-ba)=18 \end{gathered}`

so, extrema=18

Zeros: the zeros of a function is the values where the line intersect the y and x-axis, so

a) set the function to zero and solve

`\begin{gathered} x^2-9x+18=0 \\ x=\frac{-(-9)\pm\sqrt[]{(-9)^2-4\cdot1\cdot18}}{2\cdot1} \\ x=\frac{9\pm\sqrt[]{81-72}}{2} \\ x=\frac{9\pm\sqrt[]{9}}{2} \\ x=(9\pm3)/(2) \\ x_1=(9+3)/(2)=(12)/(2)=6 \\ x_2=(9-3)/(2)=(6)/(2)=3 \end{gathered}`

hence, the graph intersectsmy p the x axis at, x= 3 and x= 6

now, the y -intercetp is when the line intersecnt the y -axis, it ocurrs when x= 0, so evalute at x=0

`\begin{gathered} f(x)=x^2-9x+18 \\ f(0)=0^2-9\cdot0+18 \\ f(0)=0-0+18 \\ f(0)=18 \end{gathered}`

it means, the graph intersect the y-axis at y=18

y-intercept:18

Step 3

increasing and decreasing

a) derivate the function and equals to zero

`\begin{gathered} f(x)=x^2-9x+18 \\ y=x^2-9x+18 \\ y^(\prime)\text{ =2x-9} \\ 2x-9=0 \\ 2x=9 \\ x=(9)/(2) \\ x=4.5 \end{gathered}`

it means the inflection point is x=4.5

b) now, get the second derivate of the function to check its behavior ( if y'' is greater than zero, the function open upwards)

hence

`\begin{gathered} \text{decreasing}(-\infty,4.5) \\ \text{ increasing:(4.5,}\infty) \end{gathered}`

c) Continuity:

as the domain is all the real numbers the function is totally continuos

I hope this helps you