The first step of the problem is to find the axis of symmetry using the following formula:

\(\displaystyle x_{symmetry}=-\frac{b}{2a}\)

Where a and b are determined from the format for the equation of a parabola:

\(\displaystyle y=ax^2+bx+c\)

We can see from the equation given in the problem that a=1 and b=-3, so we can plug these values into the formula to find the axis of symmetry of our parabola:

\(\displaystyle x_{symmetry}=-\frac{-3}{2(1)}=\frac{3}{2}\)

Keep in mind that the vertex of the parabola lies directly on the axis of symmetry. That is, the x-coordinate of the axis of symmetry will be the same as that of the vertex of the parabola. Now that we know the vertex is at the same x-coordinate as the axis of symmetry, we can simply plug this value into our function to find the y-coordinate of the vertex:

\(\displaystyle y=\left(\frac{3}{2}\right)^2-3\left(\frac{3}{2}\right)-5=\frac{9}{4}-\frac{9}{2}-5=-\frac{29}{4}\)

So the vertex occurs at the point:

\(\displaystyle \left(\frac{3}{2},-\frac{29}{4}\right)\)

Rewrite the equation in standard form \(\displaystyle y=ax^2+bx+c\).

\(\displaystyle x^2=2-y\)

\(\displaystyle x^2+y=2\)

\(\displaystyle y=-x^2+0x+2\)

The vertex formula is:

\(\displaystyle x=-\frac{b}{2a}\)

Determine the necessary coefficients.

\(\displaystyle a=-1\)

\(\displaystyle b=0\)

Plug in these values to the vertex formula.

\(\displaystyle x=-\frac{b}{2a} = -\frac{0}{2(-1)} = 0\)

The axis of symmetry is \(\displaystyle x=0\).

The vertex can be thought of as the center of a parabola. Begin by finding the axis of symmetry with the following formula:

\(\displaystyle x=\frac{-b}{2a}\)

Where b and a come from the standard equation of a parabola:

\(\displaystyle y=ax^2+bx+c\)

So given our parabola

\(\displaystyle y=5x^2-25x-10\)

\(\displaystyle x=\frac{--25}{2\cdot5}=\frac{25}{10}=2.5\)

This gives us the x-coordinate of our vertex. find the y-coordinate by plugging in our x-coordinate.

\(\displaystyle y=5\cdot (2.5)^2-25\cdot (2.5)-10=-41.25\)

So our vertex is:

\(\displaystyle (2.5,-41.25)\)

The polynomial is already in \(\displaystyle y=ax^2+bx+c\) format.

To find the vertex, use the following equation:

\(\displaystyle x=-\frac{b}{2a}\)

Substitute the coefficients and solve for the vertex.

\(\displaystyle x=-\frac{3}{2(-5)}=\frac{3}{10}\)

The vertex is at \(\displaystyle x=\frac{3}{10}\).

Rewrite \(\displaystyle y=-1-2x^2\) in standard parabolic form, \(\displaystyle ax^2+bx+c\).

\(\displaystyle y=-2x^2-1\)

\(\displaystyle a=-2\)

\(\displaystyle b=0\)

\(\displaystyle c=-1\)

Write the vertex formula and substitute the values.

\(\displaystyle x=-\frac{b}{2a}=-\frac{0}{2(-2)}=0\)

The equation of the axis of symmetry is \(\displaystyle x=0\).

Substitute this value back into the original equation \(\displaystyle y=-1-2x^2\).

\(\displaystyle y=-1-2(0)^2=-1\)

The vertex is at \(\displaystyle (0,-1)\).

Find the axis of symmetry and the vertex of the parabola given by the following equation:

\(\displaystyle y=2x^2-28x+6\)

To find the axis of symmetry of a parabola in standard form, \(\displaystyle y=ax^2+bx+c\), use the following equation:

\(\displaystyle AoS=\frac{-b}{2a}\)

So...

\(\displaystyle AoS=\frac{-(-28)}{2\cdot2}=\frac{28}{4}=7\)

This means that we have an axis of symmetry at \(\displaystyle x=7\). Or, to put it more plainly, at \(\displaystyle x=7\) we could draw a vertical line which would perfectly cut our parabola in half!

So, we are halfway there, now we need the coordinates of our vertex. We already know the x-coordinate, which is 7. To find the y-coordinate, simply plug 7 into the parabola's formula and solve!

\(\displaystyle y=2(7)^2-28(7)+6=98-196+6=-92\)

This makes our vertex the point \(\displaystyle (7,-92)\)

The vertex form for a parabola is given below:

\(\displaystyle y = (x-h)^2+k\)

\(\displaystyle y=x^2 + 6x +2\) \(\displaystyle = (x^2 + 6x) +2\)

To complete the square, take the coefficient next to the x term, divide by \(\displaystyle 2\) and raise the number to the second power. In this case, \(\displaystyle (\frac{6}{2})^2 = 9\). Then take value and add it to add inside the parenthesis and subtract on the outside. 

\(\displaystyle y = (x^2 + 6x+ 9)+2-9\)

Now factor and simplify:

\(\displaystyle y = (x+3)^2 - 7\)

Fromt the values of \(\displaystyle h\) and \(\displaystyle k\), the vertex is at \(\displaystyle (-3,-7)\)

The vertex form for a parabola is given below:

\(\displaystyle y = (x-h)^2+k\)

\(\displaystyle \frac{1}{2}x^2 + 5x\)  \(\displaystyle = \frac{1}{2}(x^2 + 10x)\)

To complete the square, take the value next to the x term, divide by 2 and raise the number to the second power. In this case,\(\displaystyle (\frac{10}{2})^2 = 25\). Then take value and add it to add inside the parenthesis and subtract on the outside. 

\(\displaystyle y = \frac{1}{2}(x^2 + 10x+ 25)-\frac{25}{2}\)

Now factor and simplify:

\(\displaystyle y =\frac{1}{2} (x+5)^2 - \frac{25}{2}\)

Fromt the values of h and k, the vertex is at \(\displaystyle (-5,-\frac{25}{2})\)

The vertex form for a parabola is given below:

\(\displaystyle y = (x-h)^2+k\)

\(\displaystyle y = -2x^2+12 =\) \(\displaystyle -2(x^2-6x)\)  

To complete the square, take the value next to the x term, divide by \(\displaystyle 2\) and raise the number to the second power. In this case, \(\displaystyle (-\frac{6}{2})^2=9\). Then take value and add it to add inside the parenthesis and subtract on the outside. Remember to distribute before subtracting to the outside.

\(\displaystyle y = -2(x^2-6x+9-9) = -2(x^2-6x+9)+18\) 

Now factor and simplify:

\(\displaystyle y =-2(x-3)^2 +18\)

From the values of \(\displaystyle h\) and \(\displaystyle k\), the vertex is at \(\displaystyle (3,18)\)

The vertex form for a parabola is given below:

\(\displaystyle y = (x-h)^2+k\)

Factor the equation and transform it into the vertex form.

\(\displaystyle y= (x+2)(x-4)\) \(\displaystyle = x^2-2x-8\)

To complete the square, take the value next to the x term, divide by \(\displaystyle 2\) and raise the number to the second power. In this case, \(\displaystyle (-\frac{2}{2})^2=1\). Then take value and add it to add inside the parenthesis and subtract on the outside. 

\(\displaystyle y = (x^2-2x+1-1)-8 = (x^2-2x+1)-1-8\) 

Now factor and simplify:

\(\displaystyle y =(x-1)^2 -9\)

From the values of \(\displaystyle h\) and \(\displaystyle k\), the vertex is at \(\displaystyle (1,-9)\)