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 negative, our parabola must face down.

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

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:

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

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 positive, our parabola must face up.

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 answer is because it is the only degree-2 polynomial with a negative leading coefficient.

The answer is because it is the only degree-2 polynomial with a negative leading coefficient.