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 $\displaystyle (1,-9)$

The vertex form for a parabola is given below:

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

$\displaystyle y = -\frac{1}{3}x^2+3x+7$  $\displaystyle = -\frac{1}{3}(x^2-x )+7$

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, $\displaystyle (-\frac{1}{2})^2 =\frac{1}{4}$. Then take value and add it to add inside the parenthesis and subtract on the outside. 

$\displaystyle y = -\frac{1}{3}(x^2 - x+ \frac{1}{4}- \frac{1}{4})+7 =-\frac{1}{3}(x^2 - x+ \frac{1}{4})+7+\frac{1}{12}$

Now factor and simplify:

$\displaystyle y =-\frac{1}{3} (x-\frac{1}{2})^2 + \frac{85}{12}$

Fromt the values of $\displaystyle h$ and $\displaystyle k$, the vertex is at $\displaystyle (-\frac{1}{2},\frac{85}{12})$

The standard form for the equation of a circle with radius $\displaystyle r$, and centered at point $\displaystyle (a,b)$ is

$\displaystyle (x-a)^{2}+(y-b)^{2}=r^{2}$.

Here, $\displaystyle a=-3, b=4, r=3$, so the equation is

$\displaystyle (x+3)^{^{2}}+(y-4)^{^{2}}=9$.

Note: one way to think of this equation is to remember the Pythagorean Theorem.

If the center is at the origin then the equation is

$\displaystyle x^{2}+y^{2}=r^{2}$.

This describes a right triangle for any x and y that satisfy this equation.  Here r is the hypotenues, but when all values of x and y are used it stays the same and the points map out a circle with radius r.

The rules of graph translation apply in the same way as with any function.  That is they move the origin in the opposite direction by a and/or b.