Rewrite  in standard parabolic form, .

Write the vertex formula and substitute the values.

The equation of the axis of symmetry is .

Substitute this value back into the original equation .

The vertex is at .

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

To find the axis of symmetry of a parabola in standard form, , use the following equation:

So...

This means that we have an axis of symmetry at . Or, to put it more plainly, at  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!

This makes our vertex the point 

The vertex form for a parabola is given below:

To complete the square, take the coefficient next to the x term, divide by  and raise the number to the second power. In this case, . Then take value and add it to add inside the parenthesis and subtract on the outside. 

Now factor and simplify:

Fromt the values of  and , the vertex is at 

The vertex form for a parabola is given below:

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,. Then take value and add it to add inside the parenthesis and subtract on the outside. 

Now factor and simplify:

Fromt the values of h and k, the vertex is at 

The vertex form for a parabola is given below:

To complete the square, take the value next to the x term, divide by  and raise the number to the second power. In this case, . Then take value and add it to add inside the parenthesis and subtract on the outside. Remember to distribute before subtracting to the outside.

Now factor and simplify:

From the values of  and , the vertex is at 

The vertex form for a parabola is given below:

Factor the equation and transform it into the vertex form.

To complete the square, take the value next to the x term, divide by  and raise the number to the second power. In this case, . Then take value and add it to add inside the parenthesis and subtract on the outside. 

Now factor and simplify:

From the values of  and , the vertex is at 

The vertex form for a parabola is given below:

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, . Then take value and add it to add inside the parenthesis and subtract on the outside. 

Now factor and simplify:

Fromt the values of  and , the vertex is at 

The standard form for the equation of a circle with radius , and centered at point  is

.

Here, , so the equation is

.

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

.

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. 

The center of the circle is .

Find the horizontal distance from the center to the edge of the circle.  At the center , at the edge . The difference is .  This is the radius.

Plug these values:  into the standard form for the equation of a circle.

This gives

.

Remember that the standard form for the equation of a circle is given by the following formula:

Where the point (h,k) gives the center of the circle, and r is the radius. We can see from the form in which the equation is expressed in the problem that the only thing different with our form is that the terms on the left side of the equation are divided by 4. With some algebra, we'll multiply both sides by 4 to eliminate the 4's from the left side of equation:

Now we can see that our equation is the same as the formula for a circle in standard form, where (h,k) is (3,-2) and r=4.