Question

PLEASE HELP!!! A LOT OF POINTS AND BRAINLIEST TO CORRECT ANSWERS!!!

Answer 1

Answer:

1. $Area=16\,\pi=50.265$

2.- $\frac{b}{a} =-8$

3.  $y=\frac{1}{2} x^2+3x+\frac{5}{2}$

4.  $a+b+c=\frac{17}{4}$

5.   the vertex is located at: (3, 10)

Step-by-step explanation:

1. If we rewrite the formula of the conic given by completing squares, we can find what conic we are dealing with:

$(x^2-2x)+(y^2+6y)=6\\\,\,\,\,\,\,+1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+9\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+10\\(x-1)^2+(y+3)^2=16\\(x-1)^2+(y+3)^2=4^2$

which corresponds to a circle of radius 4, and we know what the formula is for a circle of radius R, then:

$Area=\pi\,R^2=\pi\,4^2=16\,\pi=50.265$

2.

If x=4 is the axis of symmetry of the parabola

$y=ax^2+bx+c$

then recall the formula to obtain the position of the x-value of the vertex:

$x_{vertex}=-\frac{b}{2a} \\4=-\frac{b}{2a}\\4\,(-2)=\frac{b}{a} \\\frac{b}{a} =-8$

3.  

From the graph attached, we see that the vertex of the parabola is at the point: (-3, -2) on the plane, so we can write the general formula for a parabola in vertex form:

$y-y_{vertex}=a\,(x-x_{vertex})^2\\y-(-2)=a\,(x-(-3))^2\\y+2=a(x+3)^2$

and now find the value of the parameter "a" by requesting the parabola to go through another obvious point, let's say the zero given by (-1, 0) at the crossing of the x-axis:

$y+2=a\,(x+3)^2\\0+2=a(-1+3)^2\\2=a\,2^2\\a=\frac{1}{2}$

So the equation of the parabola becomes:

$y+2=\frac{1}{2} (x+3)^2\\y+2=\frac{1}{2} (x^2+6x+9)\\y+2=\frac{1}{2} x^2+3x+\frac{9}{2} \\y=\frac{1}{2} x^2+3x+\frac{9}{2} -2\\y=\frac{1}{2} x^2+3x+\frac{5}{2}$

4.

From the location of the focus of the parabola as (4, 3) and the directrix as y=1,  we conclude that we have a parabola with dominant vertical axis of symmetry, displaced from the origin of coordinates, and responding to the following type of formula:

$(x-h)^2=4\,p\,(y-k)$

with focus at: $(h,k+p)$

directrix given by the horizontal line $y=k-p$

and symmetry axis given by the vertical line $x=h$

Since we are given that the focus is at (4, 3), we know that $h=4$, and that $k+p=3$

Now given that the directrix is: y = 1, then:

$y=k-p\\1=k-p$

Now combining both equations with these unknowns:

$k+p=3\\k=3-p$

$1=k-p\\k=1+p$

then :

$1+p=3-p\\2p=3-1\\2p=2\\p=1$

and we now can solve for k:

$k=1+p=1+1=2$

Then we have the three parameters needed to write the equation for this parabola:

$(x-h)^2=4\,p\,(y-k)\\(x-4)^2=4\,(1)\,(y-2)\\x^2-8x+16=4y-8\\4y=x^2-8x+16+8\\4y=x^2-8x+24\\y=\frac{1}{4} x^2-2x+6$

therefore: $a=\frac{1}{4} , \,\,\,b=-2,\,\,and\,\,\,c=6$

Then $a+b+c=\frac{17}{4}$

5.

The vertex of a parabola can easily found because they give you the roots of the quadratic function, which are located equidistant from the symmetry axis. So we know that is one root is at $x=3+\sqrt{5}$and the other root is at $x=3-\sqrt{5}$

then the x position of the vertex must be located at x = 3 (equidistant from and in the middle of both solutions. Then we can use the formula for the x of the vertex to find b:

$x_{vertex}=-\frac{b}{2a}\\3=-\frac{b}{2\,(-2)}\\ b=12$

Now, all we need is to find c, which we can do by using the rest of the quadratic formula for the solutions $x=3+\sqrt{5}$  and $x=3-\sqrt{5}$ :

$x=-\frac{b}{2a} +/-\frac{\sqrt{b^2-4\,a\,c} }{2\,a}$

Therefore the amount $\frac{\sqrt{b^2-4\,a\,c} }{2\,a}$,  should give us $\sqrt{5}$

which means that:

$\sqrt{5}=\frac{\sqrt{b^2-4\,a\,c} }{2\,a} \\5=\frac{b^2-4ac}{4 a^2} \\5\,(4\,(-2)^2)=(12)^2-4\,(-2)\,c\\80=144+8\,c\\8\,c=80-144\\8\,c=-64\\c=-8$

Ten the quadratic expression is:

$y=-2x^2+12\,x-8$

and the y value for the vertex is:

$y=-2(3)^2+12\,(3)-8=-18+36-8=10$

so the vertex is located at: (3, 10)