Which of the following could be the graph of f(x)?

Begin by realizing this must be a downward facing parabola with its vertex at (0,3)

We know this because of the negative sign in front of the 5, and by the constant term of 3 on the end.

This narrows our options down to 2. One is much narrower than the other, although it may seem counterintuitive, the narrower one is what we need. This is because for every increase in x, we get a corresponding increase of times 5 in y. This translates to a graph that will get to higher values of y faster than a basic parabola. So, we need the graph below. to further confirm, try to find f(1)

So, the point (1,-2) must be on the graph, which means we must have:

The coefficient of the squared term tells us whether the parabola faces up or down. Parabolas in general, as in the parent function, are in the shape of a U. In the equation given, the coefficient of the squared term is . Generally, if the coefficient of the squared term is positive, the parabola faces up. If the coefficient is negative, the parabola faces down. Since  is negative, our parabola must face down. 

The coefficient of the squared term tells us whether the parabola faces up or down. Parabolas in general, as in the parent function, are in the shape of a U. In the equation given, the coefficient of the squared term is . Generally, if the coefficient of the squared term is positive, the parabola faces up. If the coefficient is negative, the parabola faces down. Since  is positive, our parabola must face up. 

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

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

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:

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:

So the vertex occurs at the point:

Rewrite the equation in standard form .

The vertex formula is:

Determine the necessary coefficients.

Plug in these values to the vertex formula.

The axis of symmetry is .

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

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

So given our parabola

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

So our vertex is:

The polynomial is already in  format.

To find the vertex, use the following equation:

Substitute the coefficients and solve for the vertex.

The vertex is at .

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