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ICE Princess25 [194]
3 years ago
12

Whats the vertex of the graph y=-x^2

Mathematics
1 answer:
yawa3891 [41]3 years ago
6 0
(0,0) you could use x=-b/2a but since your b term is nonexistant the whole term goes to 0 for the x value. Plug x=0 into the original equation which gives you another 0. ezpz
You might be interested in
An object is launched from a platform. Its height (in meters), xxx seconds after the launch, is modeled by: h(x)=-5x^2+20x+60h(x
EastWind [94]
It takes 6 seconds for it to hit the ground.

0 = -5x²+20x+60

We can solve this by factoring.  First factor out the GCF, -5:

0 = -5(x²-4x-12)

Now we want factors of -12 that sum to -4.  -6(2) = -12 an -6+2 = -4:
0 = -5(x-6)(x+2)

Using the zero product property, we know that either x-6=0 or x+2=0; this gives us the answers x=6 or x=-2.  Since we cannot have negative time, x=6.
3 0
3 years ago
Read 2 more answers
HELP PLEASEEE
ivanzaharov [21]

Answer:

1. 4

2. 16

3. 4/3

4. 64/3

Step-by-step explanation:

3 0
3 years ago
Explain how you got the answer
Gnoma [55]

Answer:

\huge\boxed{Center = (3,3),Radius = 2\sqrt{15} }

Step-by-step explanation:

<u><em>Given equation is</em></u>

x^2 + y^2 -6x-6y -42 = 0

Adding 42 to both sides

x^2 + y^2 -6x-6y = 42\\

Completing squares

x^2 -6x+y^2 -6x = 42\\(x)^2 - 2(x)(3) +y^2 - 2(y)(3) = 42

Adding (3)² => 9 and (3)² => 9 to both sides

(x-3)^2+(y-3)^2 = 42+9+9\\(x-3)^2 + (y-3)^2 = 60\\(x-3)^2  (y-3)^2 = (2{\sqrt{15})^2}

Comparing it with (x-h)^2+(y-k)^2 = r^2 where Center = (h,k) and Radius = r

We get:

Center = (3,3)

Radius = 2\sqrt{15}

7 0
3 years ago
Help please i will mark brainlyy
skelet666 [1.2K]

Answer:

x =(-4+√88)/12=-1/3+1/6√ 22 = 0.448

x =(-4-√88)/12=-1/3-1/6√ 22 = -1.115

Step-by-step explanation:

x =(-4-√88)/12=-1/3-1/6√ 22 = -1.115

x =(-4+√88)/12=-1/3+1/6√ 22 = 0.448

Reformatting the input :

Changes made to your input should not affect the solution:

(1): "x2"   was replaced by   "x^2".  

Step by step solution :

Step  1  :

Equation at the end of step  1  :

 ((2•3x2) +  4x) -  3  = 0  

Step  2  :

Trying to factor by splitting the middle term

2.1     Factoring  6x2+4x-3  

The first term is,  6x2  its coefficient is  6 .

The middle term is,  +4x  its coefficient is  4 .

The last term, "the constant", is  -3  

Step-1 : Multiply the coefficient of the first term by the constant   6 • -3 = -18  

Step-2 : Find two factors of  -18  whose sum equals the coefficient of the middle term, which is   4 .

     -18    +    1    =    -17  

     -9    +    2    =    -7  

     -6    +    3    =    -3  

     -3    +    6    =    3  

     -2    +    9    =    7  

     -1    +    18    =    17  

Observation : No two such factors can be found !!

Conclusion : Trinomial can not be factored

Equation at the end of step  2  :

 6x2 + 4x - 3  = 0  

Step  3  :

Parabola, Finding the Vertex :

3.1      Find the Vertex of   y = 6x2+4x-3

Parabolas have a highest or a lowest point called the Vertex .   Our parabola opens up and accordingly has a lowest point (AKA absolute minimum) .   We know this even before plotting  "y"  because the coefficient of the first term, 6 , is positive (greater than zero).  

Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two  x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions.  

Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex.  

For any parabola,Ax2+Bx+C,the  x -coordinate of the vertex is given by  -B/(2A) . In our case the  x  coordinate is  -0.3333  

Plugging into the parabola formula  -0.3333  for  x  we can calculate the  y -coordinate :  

 y = 6.0 * -0.33 * -0.33 + 4.0 * -0.33 - 3.0

or   y = -3.667

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = 6x2+4x-3

Axis of Symmetry (dashed)  {x}={-0.33}  

Vertex at  {x,y} = {-0.33,-3.67}  

x -Intercepts (Roots) :

Root 1 at  {x,y} = {-1.12, 0.00}  

Root 2 at  {x,y} = { 0.45, 0.00}  

Solve Quadratic Equation by Completing The Square

3.2     Solving   6x2+4x-3 = 0 by Completing The Square .

Divide both sides of the equation by  6  to have 1 as the coefficient of the first term :

  x2+(2/3)x-(1/2) = 0

Add  1/2  to both side of the equation :

  x2+(2/3)x = 1/2

Now the clever bit: Take the coefficient of  x , which is  2/3 , divide by two, giving  1/3 , and finally square it giving  1/9  

Add  1/9  to both sides of the equation :

 On the right hand side we have :

  1/2  +  1/9   The common denominator of the two fractions is  18   Adding  (9/18)+(2/18)  gives  11/18  

 So adding to both sides we finally get :

  x2+(2/3)x+(1/9) = 11/18

Adding  1/9  has completed the left hand side into a perfect square :

  x2+(2/3)x+(1/9)  =

  (x+(1/3)) • (x+(1/3))  =

 (x+(1/3))2

Things which are equal to the same thing are also equal to one another. Since

  x2+(2/3)x+(1/9) = 11/18 and

  x2+(2/3)x+(1/9) = (x+(1/3))2

then, according to the law of transitivity,

  (x+(1/3))2 = 11/18

We'll refer to this Equation as  Eq. #3.2.1  

The Square Root Principle says that When two things are equal, their square roots are equal.

Note that the square root of

  (x+(1/3))2   is

  (x+(1/3))2/2 =

 (x+(1/3))1 =

  x+(1/3)

Now, applying the Square Root Principle to  Eq. #3.2.1  we get:

  x+(1/3) = √ 11/18

Subtract  1/3  from both sides to obtain:

  x = -1/3 + √ 11/18

Since a square root has two values, one positive and the other negative

  x2 + (2/3)x - (1/2) = 0

  has two solutions:

 x = -1/3 + √ 11/18

  or

 x = -1/3 - √ 11/18

Note that  √ 11/18 can be written as

 √ 11  / √ 18  

Solve Quadratic Equation using the Quadratic Formula

3.3     Solving    6x2+4x-3 = 0 by the Quadratic Formula .

According to the Quadratic Formula,  x  , the solution for   Ax2+Bx+C  = 0  , where  A, B  and  C  are numbers, often called coefficients, is given by :

                                     

           - B  ±  √ B2-4AC

 x =   ————————

                     2A

 In our case,  A   =     6

                     B   =    4

                     C   =   -3

Accordingly,  B2  -  4AC   =

                    16 - (-72) =

                    88

Applying the quadratic formula :

              -4 ± √ 88

  x  =    —————

                   12

Can  √ 88 be simplified ?

Yes!   The prime factorization of  88   is

  2•2•2•11  

To be able to remove something from under the radical, there have to be  2  instances of it (because we are taking a square i.e. second root).

√ 88   =  √ 2•2•2•11   =

               ±  2 • √ 22

 √ 22   , rounded to 4 decimal digits, is   4.6904

So now we are looking at:

          x  =  ( -4 ± 2 •  4.690 ) / 12

Two real solutions:

x =(-4+√88)/12=-1/3+1/6√ 22 = 0.448

or:

x =(-4-√88)/12=-1/3-1/6√ 22 = -1.115

5 0
3 years ago
Read 2 more answers
What set of values is apart of the solution set to the inequality 3.5p+14&gt;7p
aalyn [17]

Answer:

q

Step-by-step explanation:

5 0
3 years ago
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