Question

1-Use the vertex and intercepts to sketch the graph of the quadratic function. Give the equation of the​ parabola's axis of symmetry. Use the graph to determine the domain and range of the function.

f(x)=(x+4)^2−9

2-

A. Graph

f left parenthesis x right parenthesis equals negative x squared minus 4 xf(x)=−x2−4x

by determining whether its graph opens up or down and by finding its​ vertex, axis of​ symmetry, y-intercept, and​ x-intercepts, if any.

B. Determine the domain and the range of the function.

C. Determine where the function is increasing and where it is decreasing

Answer 1

1)
`x=-4`
`D=(-\infty,\infty)`
`R=[-9,\infty)` 1st graph below x-intercepts S={-7,-1}

2) Opens Down Vertex: (-2,4) Axis of Symmetry
`x=-2` x-intercept
`S=\left \{ 0,-4 \right \} <u>y-intercept: (0,0)</u>`

B Domain:
`D=(-\infty,\infty)`

Range
`R=(-\infty,4]`

C This function increases from
`(-\infty,-2)` And decreases from
`(-2,\infty)`

Explanation:

1) The vertex of the parabola is found when we rewrite the common formula:

`f(x)=ax^(2)+bx+c`

Into this way:

`f(x)=a(x-h)^(2)+k` The Vertex is found by:

`\\h=(-b)/(2a);k=(-\Delta )/(4a)\\f(x)=y=x^(2)+8x+16-9\Rightarrow y=x^(2)+8x+7\Rightarrow h=(-8)/(2)\Rightarrow h=-4\Rightarrow k=(-\Delta )/(4a)\Rightarrow \:k=-(8^(2)-(4*1*7))/(4*1)\Rightarrow k=-9\\ (h,k)\Rightarrow (-4,-9)`

That's why we could rewrite the trinomial as this:

`f(x)=(x+4)^2-9\\`

Give the equation of the​ parabola's axis of symmetry, this is given by tracing a vertical line through the parabola vertex.

`x=-4`

The intercepts are the roots/zeros:

`y=x^(2)+8x+7\Rightarrow y=(x+7)(x+1) \Rightarrow x'=-7,\:x''=-1 \:S=\left \{ -7,-1 \right \}`

Domain:

Since the function has no restrictions therefore it is continuous and defined for any value of x ∈ Real Set

`D=(-\infty,\infty)`

Range

As the minimum point -9 is lowest y-coordinate the Range includes this value up to infinite values

`R=[-9,\infty)`

(First Graph)

2)
`f(x)=-x^(2)-4x`

As the parameter a <0 then the graph opens down.

Vertex:

`h=(-b)/(2a);k=(-\Delta )/(4a)\Rightarrow h=-\left ( (-4)/(2(-1)) \right ); k=-\left ( ((-4)^(2)-4(-1)(0))/(4(-1)) \right )\Rightarrow (-2,4)`

Axis of Symmetry

`x=-2`

x-intercept

`\\y=-x^(2)-4x\Rightarrow 0=-x^(2)-4x\Rightarrow x^(2)+4x=0\Rightarrow x(x+4)=0\Rightarrow S=\left \{ 0,-4 \right \}`

y-intercept

c=0 then (0,0).

B Domain:

Similarly, since the function has no restrictions therefore it is continuous it is defined for any value of x ∈ Real Set

`D=(-\infty,\infty)`

Range

`R=(-\infty,4]`

C

This function increases from
`(-\infty,-2)` Or we can represent this interval like this:

And decreases from
`(-2,\infty)`