We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

PSAT Math : How to use the quadratic function

Study concepts, example questions & explanations for PSAT Math

Create An Account Create Tests & Flashcards

All PSAT Math Resources

10 Diagnostic Tests 421 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 :How To Use The Quadratic Function

If x²+ 2x - 1 = 7, which answers for x are correct?

Possible Answers:

x = -3, x = 4

x = 8, x = 0

x = -4, x = -2

x = -4, x = 2

x = -5, x = 1

Correct answer:

x = -4, x = 2

Explanation:

x²+ 2x - 1 = 7

x²+ 2x - 8 = 0

(x + 4) (x - 2) = 0

x = -4, x = 2

Report an Error

Example Question #2 :How To Use The Quadratic Function

Which of the following quadratic equations has a vertex located at $\dpi{100} (3,4)$ ?

Possible Answers:

$f(x)=-2x^2-12x+4$

$f(x)=-2x^2-12x+58$

$f(x)=-2x^2+8x-2$

$f(x)=-2x^2+12x-14$

$f(x)=-2x^2+12x-12$

Correct answer:

$f(x)=-2x^2+12x-14$

Explanation:

The vertex form of a parabola is given by the equation:

$f(x)=a(x-h)^2 +k$ , where the point $\dpi{100} (h,k)$ is the vertex, and $\dpi{100} a$ is a constant.

We are told that the vertex must occur at $\dpi{100} (3,4)$ , so let's plug this information into the vertex form of the equation. $\dpi{100} h$ will be 3, and $\dpi{100} k$ will be 4.

$f(x)=a(x-3)^2 +4$

Let's now expand $(x-3)^2$ 通过使用the FOIL method, which requires us to multiply the first, inner, outer, and last terms together before adding them all together.

$(x-3)^2 = (x-3)(x-3)=x^2-3x-3x+9=x^2-6x+9$

We can replace $(x-3)^2$ with $x^2-6x+9$ .

$f(x)=a(x-3)^2+4=a(x^2-6x+9)+4$

Next, distribute the $\dpi{100} a$ .

$a(x^2-6x+9)+4 = ax^2 -6ax+9a+4$

Notice that in all of our answer choices, the first term is $-2x^2$ . If we let $\dpi{100} a=-2$ , then we would have $-2x^2$ 在我们的方程。让我们看看会发生什么bstitute $\dpi{100} -2$ for $\dpi{100} a$ .

$f(x)=ax^2-6ax+9a+4=(-2)x^2-6(-2)x+9(-2)+4$

$=-2x^2+12x-18+4$

Report an Error

Example Question #1 :How To Use The Quadratic Function

Use the quadratic equation to solve for $\small x$ .

$\small \small \small \small x^{2} + 3x - 3 = 0$

Possible Answers:

$\small \small \small x = \small \frac{-3\pm \sqrt{3}}{2}$

None of the other answers

$\small \small \small \small \small \small x = \small \frac{3\pm \sqrt{3}}{2}$

$\small \small \small \small x = \small \frac{3\pm \sqrt{21}}{2}$

$\small \small x = \small \frac{-3\pm \sqrt{21}}{2}$

Correct answer:

$\small \small x = \small \frac{-3\pm \sqrt{21}}{2}$

Explanation:

We take a polynomial in the form $\small ax^2 + bx +c = 0$

and enter the corresponding coefficients into the quadratic equation.

$\small \small \frac{-b\pm \sqrt{b^2 - 4ac}}{2a} = x$ . We normally expect to have two answers given by the sign $\small \pm$ .

So,

$\small \small \small \small x^{2} + 3x - 3 = 0$

$\small \small \small \small \frac{-(3)\pm \sqrt{(3)^2 - 4(1)(-3)}}{2(1)} = x$

$\small \small \small \small \frac{-3\pm \sqrt{9 + 12}}{2} = x$

$\small \small \small \small \small \frac{-3\pm \sqrt{21}}{2} = x$

Report an Error

Example Question #2 :How To Use The Quadratic Function

Define functionas follows:

$f(x) = x^{2} - 10x + 7$

Given that f (a) = f(5) and $a \ne 5$ , evaluate.

Possible Answers:

a = 0

a = 12

a = 2

No such value exists.

a= -5

Correct answer:

No such value exists.

Explanation:

$f(x) = x^{2} - 10x + 7$

$f(5) = 5^{2} - 10 \cdot 5 + 7$

= 25 - 50 + 7

= -18

We solve forin the equation

f (a) = f(5)

f(a) = -18

$a^{2} - 10a+ 7 = -18$

$a^{2} - 10a+ 25= 0$

$(a-5)^{2} = 0$

a-5 = 0

a=5

This is the only solution. Since it is established thatis not equal to 5, the correct response is that no such value exists.

Report an Error

Example Question #3 :How To Use The Quadratic Function

Define functionas follows:

$f(x) = x^{2} - 6x + 5$

Given that f (a) = f(9) and $a \ne 9$ , evaluate.

Possible Answers:

a = -3

a = 1

No such value exists.

a = 14

a = 4

Correct answer:

a = -3

Explanation:

$f(x) = x^{2} - 6x + 5$

$f(9) = 9^{2} - 6 \cdot 9 + 5$

= 81 - 54 + 5

= 32

Solve forin the equation

f (a) = f(9)

f(a) = 32

$a^{2} - 6a + 5 = 32$

$a^{2} - 6a -27 = 0$

(a-9)(a+3) = 0

Either a- 9 = 0 , in which case a = 9 , which is already established to be untrue, or a + 3 = 0 , in which case a = -3 . This is the correct response.

Report an Error

Example Question #4 :How To Use The Quadratic Function

A pitcher standing on top on a 600-foot building throws a baseball upward at an initial speed of 90 feet per second. The heightof the ball at a given timecan be modeled by the function

$h\left ( t\right )= -16t^{2}+ 90 t + 600$

How high does the ball get? (Nearest foot)

Possible Answers:

$674\textup{ ft}$

$727\textup{ ft}$

$1,194 \textup{ ft}$

$977\textup{ ft}$

$708\textup{ ft}$

Correct answer:

$727\textup{ ft}$

Explanation:

This can be solved by first finding the first coordinate (value) of the vertex of the parabola representing the function.

The-coordinate of the vertex is $-\frac{b}{2a}$ , where a = -16, b = 90 ;

$-\frac{b}{2a} = -\frac{90}{2(-16) } = 2.8$

The ball takes 2.8 seconds to reach its peak. The height at that time is h (2.8) , which is evaluated using substitution:

$h\left ( t\right )= -16t^{2}+ 90 t + 600$

$h\left ( 2.8 \right )= -16\cdot 2.8^{2}+ 90 \cdot 2.8+ 600$

$= -16\cdot 7.84+ 90 \cdot 2.8+ 600$

= -125.44+ 252+ 600

= 726.56

making the correct response 727 feet.

Report an Error

Example Question #5 :How To Use The Quadratic Function

A pitcher standing on top on a 600-foot-high building throws a baseball upward at an initial speed of 90 feet per second. The heightof the ball at a given timecan be modeled by the function

$h\left ( t\right )= -16t^{2}+ 90 t + 600$

How long does it take for the ball to hit the ground? (Nearest tenth of a second)

Possible Answers:

$3.9\textup{ sec}$

$5.6\textup{ sec}$

$9.6\textup{ sec}$

$2.8 \textup{ sec}$

$6.7\textup{ sec}$

Correct answer:

$9.6\textup{ sec}$

Explanation:

Set the height function equal to 0:

$h\left ( t\right )=0$

$-16t^{2}+ 90 t + 600 = 0$

Set a = -16, b = 90, c = 600 in the quadratic formula

$t = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

$t = \frac{-90 \pm \sqrt{90^{2}-4(-16)(600)}}{2 (-16)}$

$t = \frac{-90 \pm \sqrt{8,100+38,400}}{-32}$

$t = \frac{-90 \pm \sqrt{46,500}}{-32}$

$t \approx \frac{-90 \pm 215.64}{-32}$

Evaluate separately, and select thepositivevalue.

$t \approx \frac{-90 +215.64}{-32} \approx \frac{125.64}{-32} \approx -3.9$

which is thrown out.

$t \approx \frac{-90 -215.64}{-32} \approx \frac{-305.64}{-32} \approx 9.6$

which is positive and is the result we keep.

The correct response is 9.6 seconds.

Report an Error

Example Question #6 :How To Use The Quadratic Function

A pitcher standing on top on a 600-foot building throws a baseball upward at an initial speed of 90 feet per second. The heightof the ball at a given timecan be modeled by the function

$h\left ( t\right )= -16t^{2}+ 90 t + 600$

How long does it take for the ball to return to the level at which it started? (Nearest tenth of a second)

Possible Answers:

$9.6\textup{ sec}$

$2.8 \textup{ sec}$

$6.7\textup{ sec}$

$5.6\textup{ sec}$

$3.9\textup{ sec}$

Correct answer:

$5.6\textup{ sec}$

Explanation:

这个问题本质上是要求是e value of timewhen

h(t) = 600 .

$-16t^{2}+ 90 t + 600 = 600$

$-16t^{2}+ 90 t = 0$

$-2 t \left (8t - 45 \right ) = 0$

Either t = 0 - but this simply reflects that theinitialheight was 600 feet - or:

8t-45 = 0

8t=45

$t = 45 \div 8 \approx 5.6$ .

The ball returns to a height of 600 feet after 5.6 seconds.

Report an Error

Example Question #7 :How To Use The Quadratic Function

A pitcher standing on top on a 600-foot building throws a baseball upward at an initial speed of 90 feet per second. The heightof the ball at a given timecan be modeled by the function

$h\left ( t\right )= -16t^{2}+ 90 t + 600$

How long does it take for the ball to reach its peak? (Nearest tenth of a second)

Possible Answers:

$3.9\textup{ sec}$

$9.6\textup{ sec}$

$6.7\textup{ sec}$

$5.6\textup{ sec}$

$2.8 \textup{ sec}$

Correct answer:

$2.8 \textup{ sec}$

Explanation:

The time at which the ball reaches its peak can be found by finding the-coordinate of the vertex of the parabola representing the function.

The-coordinate of the vertex is $-\frac{b}{2a}$ , where a = -16, b = 90 ;

$-\frac{b}{2a} = -\frac{90}{2(-16) } = 2.8$

The correct response is 2.8 seconds.

Report an Error

View PSAT Mathematics Tutors

Vishu
Certified Tutor

University of Toronto, Bachelor of Science, Computer Engineering, General.

View PSAT Mathematics Tutors

Ari
Certified Tutor

Georgetown University, Bachelor of Science, International Relations.

View PSAT Mathematics Tutors

Hadassah
Certified Tutor

Oakwood University, Bachelor of Science, Mathematics Teacher Education. University of Phoenix-Online Campus, Masters in Educa...

All PSAT Math Resources

10 Diagnostic Tests 421 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

ACT Tutors in Los Angeles,French Tutors in Dallas Fort Worth,SSAT Tutors in Los Angeles,Statistics Tutors in Chicago,ISEE Tutors in Chicago,GRE Tutors in Seattle,ISEE Tutors in Washington DC,GRE Tutors in Boston,French Tutors in Philadelphia,GRE Tutors in New York City

Popular Courses & Classes

LSAT Courses & Classes in Boston,SSAT Courses & Classes in Miami,在凤凰城坐课程&类,ISEE Courses & Classes in Washington DC,ISEE Courses & Classes in Houston,SSAT Courses & Classes in Philadelphia,LSAT Courses & Classes in Philadelphia,GRE Courses & Classes in Chicago,GRE Courses & Classes in Washington DC,LSAT Courses & Classes in Seattle

Popular Test Prep

SSAT Test Prep in Seattle,LSAT Test Prep in San Diego,ISEE Test Prep in Phoenix,GRE Test Prep in Chicago,SSAT Test Prep in Atlanta,SSAT Test Prep in Los Angeles,MCAT Test Prep in Phoenix,MCAT Test Prep in New York City,GRE Test Prep in Los Angeles,ISEE Test Prep in Boston

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

*Full legal name

Company name

*Phone number

*Email address

Address

*Address1

Address2

*City

*State

*Zipcode

*Country

Complaint Details

*URL of copyrighted work (where your original material is located)

*Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

*Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

PSAT Math : How to use the quadratic function

Study concepts, example questions & explanations for PSAT Math

All PSAT Math Resources

Example Questions

Example Question #1 :How To Use The Quadratic Function

Example Question #2 :How To Use The Quadratic Function

Example Question #1 :How To Use The Quadratic Function

Example Question #2 :How To Use The Quadratic Function

Example Question #3 :How To Use The Quadratic Function

Example Question #4 :How To Use The Quadratic Function

Example Question #5 :How To Use The Quadratic Function

Example Question #6 :How To Use The Quadratic Function

Example Question #7 :How To Use The Quadratic Function

All PSAT Math Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Find the Best Tutors

Email address:
Your name:
Feedback: