PSAT Math : How to use the quadratic function

Study concepts, example questions & explanations for PSAT Math

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Example Questions

Example Question #208 : Algebraic Functions

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

Possible Answers:

x = -5, x = 1

x = 8, x = 0

x = -4, x = -2

x = -4, x = 2

x = -3, x = 4

Correct answer:

x = -4, x = 2

Explanation:

x + 2x - 1 = 7

x + 2x - 8 = 0

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

x = -4, x = 2

Example Question #111 : Algebraic Functions

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

Possible Answers:

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

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

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

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

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

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 by using 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 in our equation. Let's see what happens when we substitute \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

Example Question #1 : How To Use The Quadratic Function

Use the quadratic equation to solve for .

 

Possible Answers:

None of the other answers

Correct answer:

Explanation:

We take a polynomial in the form 

and enter the corresponding coefficients into the quadratic equation.

.  We normally expect to have two answers given by the sign .

So,

 

 

Example Question #2 : How To Use The Quadratic Function

Define function  as follows:

Given that  and , evaluate .

Possible Answers:

No such value exists.

Correct answer:

No such value exists.

Explanation:

We solve for  in the equation

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

Example Question #3 : How To Use The Quadratic Function

Define function  as follows:

Given that  and , evaluate .

Possible Answers:

No such value exists.

Correct answer:

Explanation:

Solve for  in the equation

Either , in which case , which is already established to be untrue, or , in which case . This is the correct response.

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 height  of the ball at a given time  can be modeled by the function

How high does the ball get? (Nearest foot)

Possible Answers:

Correct answer:

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 , where ;

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

making the correct response 727 feet.

 

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 height  of the ball at a given time  can be modeled by the function

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

Possible Answers:

Correct answer:

Explanation:

Set the height function equal to 0:

Set  in the quadratic formula

Evaluate separately, and select the positive value.

which is thrown out.

which is positive and is the result we keep.

The correct response is 9.6 seconds.

Example Question #5 : 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 height  of the ball at a given time  can be modeled by the function

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

Possible Answers:

Correct answer:

Explanation:

The question is essentially asking for the positive value of time  when 

.

Either  - but this simply reflects that the initial height was 600 feet - or:

.

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

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 height  of the ball at a given time  can be modeled by the function

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

Possible Answers:

Correct answer:

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 , where ;

The correct response is 2.8 seconds.

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