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

Answer:

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 $S=\left \{ 0,-4 \right \} y-intercept: (0,0)$

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

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

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

Step-by-step 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:

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

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:

$h=\frac{-b}{2a};k=\frac{-\Delta }{4a}\Rightarrow h=-\left ( \frac{-4}{2(-1)} \right ); k=-\left ( \frac{(-4)^{2}-4(-1)(0)}{4(-1)} \right )\Rightarrow (-2,4)$

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 $(-2,\infty)$