SAT Mathematics : Graphing Quadratics & Polynomials

Example Questions

Example Question #1 : Graphical Representation Of Functions

is a nonzero constant in the quadratic equation above. When graphed in the -plane, the equation is a parabola with a vertex at point . What is the value of ?

Explanation:

There are two ways to solve this problem. For one, you might see that the problem involves a vertex and decide to put the equation in vertex form, , where the vertex is at point . To employ vertex form, recognize that you will need to get the quadratic to be a perfect square, using the  term to balance.  Here if you were to multiply out the quadratic in the given equation, you'd have:

Note that the only perfect square quadratic with  as the middle term is , which factors to the perfect square .  So in order to put this equation in vertex form, you need to convert  into , where  equals out the step you took to turn  within the parentheses to . To do that step, you added  (remember that the entire parenthetical is multiplied by ). So you need to subtract  on the outside of the parentheses to keep the equation balanced. That means that you now have:

, or when factored directly to vertex form

This means that the vertex is at , or more directly that . So the correct answer is .

You could also use the line of symmetry to solve here. You know that the zeros of the given equation, , will be at  and , as those are the points at which each parenthetical term equals zero, so  would in turn equal zero. Because parabolas are symmetrical, that tells you that the vertex will lie halfway between those two points, at . If you then plug in the -coordinate of the vertex, , you'll get the corresponding -coordinate:

.

Example Question #2 : Graphical Representation Of Functions

The parabola pictured above is defined by the equation  and has a vertex at point . Which of the following is true of a parabola defined by the equation  ?

The parabola opens downward and has a vertex at point

The parabola opens upward and has a vertex at point

The parabola opens upward and has a vertex at point

The parabola opens downward and has a vertex at point

The parabola opens downward and has a vertex at point

Explanation:

Vertex form of a parabola is , where the point  is the vertex of the parabola, and the sign of  determines which way the parabola opens. A negative  opens downward, and a positive  opens upward. Here the given parabola has established that the value of  without the negative sign would point upward, so multiplying  by  will flip the parabola to point downward. And in Vertex Form,  and  are  and , so the vertex will be at .

Example Question #3 : Graphical Representation Of Functions

Complete the square to calculate the maximum or minimum point of the given function.

Explanation:

Completing the square method uses the concept of perfect squares. Recall that a perfect square is in the form,

where when multiplied out,

the middle term coefficient, when divided by two and squared, results in the coefficient of the last term.

Complete the square for this particular function is as follows.

First identify the middle term coefficient.

Now divide the middle term coefficient by two.

From here write the function with the perfect square. Remember when adding the new squared term, add it to both sides to keep the equation balanced.

When simplified the new function is,

Since the  term is positive, the parabola will be opening up. This means that the function has a minimum value at the vertex. To find the  value of the vertex set the inside portion of the binomial equal to zero and solve.

From here, substitute the the  value into the original function.

Therefore the minimum value occurs at the point .

Example Question #4 : Graphical Representation Of Functions

Complete the square to calculate the maximum or minimum point of the given function.

Explanation:

Completing the square method uses the concept of perfect squares. Recall that a perfect square is in the form,

where when multiplied out,

the middle term coefficient, when divided by two and squared, results in the coefficient of the last term.

Complete the square for this particular function is as follows.

First identify the middle term coefficient.

Now divide the middle term coefficient by two.

From here write the function with the perfect square.

When simplified the new function is,

Since the  term is positive, the parabola will be opening up. This means that the function has a minimum value at the vertex. To find the  value of the vertex set the inside portion of the binomial equal to zero and solve.

From here, substitute the the  value into the original function.

Therefore the minimum value occurs at the point .

Example Question #5 : Graphical Representation Of Functions

Complete the square to calculate the maximum or minimum point of the given function.

Explanation:

Completing the square method uses the concept of perfect squares. Recall that a perfect square is in the form,

where when multiplied out,

the middle term coefficient, when divided by two and squared, results in the coefficient of the last term.

Complete the square for this particular function is as follows.

First factor out a negative one.

Now identify the middle term coefficient.

Now divide the middle term coefficient by two.

From here write the function with the perfect square.

When simplified the new function is,

Since the  term is negative, the parabola will be opening down. This means that the function has a maximum value at the vertex. To find the  value of the vertex set the inside portion of the binomial equal to zero and solve.

From here, substitute the the  value into the original function.

Therefore the maximum value occurs at the point .

Example Question #6 : Graphical Representation Of Functions

In which quadrant of the xy-plane does the vertex of the following function lie?

The vertex lies on one of the axes, not in a quadrant.

Explanation:

Finding the vertex of a quadratic function is best accomplished by finding  and  in the equation in vertex form where ,  or by leveraging the formulas for  and . So your goal should be to take the given equation and put it in that form. You start with:

Start by factoring out the common 3 to see that:

Recognize that the perfect square for the terms  will be  and then adjust the remaining term to keep equivalence:

Lastly, you need to get the – 5 out of the parentheses so that you are in the required form . With this last step you see that:

So,  and

Because  and , the point  is meaning that the vertex lies in Quadrant III.

Note: You could also find the vertex by simply leveraging the two formulas below:

Since , , , we can plug into the equation to see that

Thus the vertex is .

Example Question #7 : Graphical Representation Of Functions

How many distinct roots does the polynomial  have?

1

3

2

4

3

Explanation:

This equation isn’t at all impossible to factor, but you may not notice that at first glance (or on a future problem you may try to factor and get stuck). If you see this problem on the calculator section, it’s likely most efficient to simply graph it on your calculator and count the number of times that the graph touches the y-axis. Your graph will look like:

As you can see, the graph crosses the y-axis twice and touches it a third time (at the origin, or point (0, 0)), so the answer is 3. Of course, you could factor this equation by factoring out like terms. Factoring out the common x2 term creates: , which can then be factored to Since your goal is for y to equal 0 (the definition of “a zero” in a function or a “root” in an equation), you can set that equal to 0: This means that the roots are 0, −6, and 3, for a total of 3 roots.

Example Question #8 : Graphical Representation Of Functions

Which of the following represents the maximum or minimum point of the given function?

(-1, 5)

(1, 5)

(-1, -5)

(1, -5)

(-1, -5)

Explanation:

Completing the square method uses the concept of perfect squares. Recall that a perfect square is in the form,

where when multiplied out,

the middle term coefficient, when divided by two and squared, results in the coefficient of the last term.

Complete the square for this particular function is as follows.

First identify the middle term coefficient.

Middle Term Coefficient=2

Now divide the middle term coefficient by two.

From here write the function with the perfect square. Remember when adding the new squared term, add it to both sides to keep the equation balanced.

When simplified the new function is,

Since the  term is positive, the parabola will be opening up. This means that the function has a minimum value at the vertex. To find the x value of the vertex set the inside portion of the binomial equal to zero and solve.

From here, substitute the the x value into the original function.

Therefore the minimum value occurs at the point (1,5).

Example Question #9 : Graphical Representation Of Functions

Which of the following is an equivalent form of the function f above for which the minimum value of f appears as a constant or coefficient?

Explanation:

By either graphing the function, utilizing the formula for the vertex, or completing the square, we can see that the minimum value of the function is -49. In order to comply with the parameters of the question stem, this minimum must visually appear in the answer option, so

is our only feasible option. That said, we can prove this process by "completing the square."

Here, if we foil

,

we arrive at

to complete the square, we want to add, and subsequently subtract 4 as shown below

which simplifies to

and can be rewritten as

our equivalent equation in vertex form!

Example Question #10 : Graphical Representation Of Functions

Which of the following are distinct zeros does the function

I) -9

II) -4

III) 4

I and III

I only

II and III

none of the above

I and III

Explanation:

To tackle this question, we'll want to see the number sense behind factoring by inspection. In the quadratic construction

we're looking for two numbers that multiply to give us our constant, "c," and sum to give us the coefficient "b."

In this case, those two numbers are 9 and -4, meaning we can reconstruct the function to read

since we're looking for our x intercepts or "zeros," we want the case where our output or "y" = 0, so we can set the righthand side of the function equal to zero as such

in order for this equation to hold true, either of our parentheticals will need to equal zero, meaning that if

or

then x could be equal to -9 or 4. So, I and III are zeros of the function.

*Note - if given this question on a calculator-friendly section, we could graph and identify zeros as well*