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PSAT Math : How to use the quadratic function

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Example Questions

Example Question #202 : Algebraic Functions

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

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Example Question #1 : How To Use The Quadratic Function

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

Possible Answers:

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

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

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

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

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

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 $(x-3)^2$ 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 $\dpi{100} a$ .

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

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

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Example Question #3 : 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:

None of the other answers

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

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

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

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

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,

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

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

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

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

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Example Question #3 : How To Use The Quadratic Function

Define function as follows:

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

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

Possible Answers:

a = 12

a = 0

a= -5

a = 2

No such value exists.

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 for in 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 that is not equal to 5, the correct response is that no such value exists.

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Example Question #4 : How To Use The Quadratic Function

Define function as follows:

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

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

Possible Answers:

a = 1

a = -3

a = 4

No such value exists.

a = 14

Correct answer:

a = -3

Explanation:

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

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

= 81 - 54 + 5

= 32

Solve for in 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.

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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

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

How high does the ball get? (Nearest foot)

Possible Answers:

$708\textup{ ft}$

$1,194 \textup{ ft}$

$674\textup{ ft}$

$977\textup{ ft}$

$727\textup{ ft}$

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.

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Example Question #7 : 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

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

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

Possible Answers:

$2.8 \textup{ sec}$

$3.9\textup{ sec}$

$9.6\textup{ sec}$

$5.6\textup{ sec}$

$6.7\textup{ sec}$

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 the positive value.

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

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.

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Example Question #1 : 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

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

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

Possible Answers:

$2.8 \textup{ sec}$

$9.6\textup{ sec}$

$3.9\textup{ sec}$

$5.6\textup{ sec}$

$6.7\textup{ sec}$

Correct answer:

$5.6\textup{ sec}$

Explanation:

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

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 the initial height 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.

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Example Question #111 : Algebraic Functions

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

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

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

Possible Answers:

$3.9\textup{ sec}$

$2.8 \textup{ sec}$

$5.6\textup{ sec}$

$6.7\textup{ sec}$

$9.6\textup{ sec}$

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.

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PSAT Math : How to use the quadratic function

Study concepts, example questions & explanations for PSAT Math

All PSAT Math Resources

Example Questions

Example Question #202 : Algebraic Functions

Example Question #1 : How To Use The Quadratic Function

Example Question #3 : 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 #4 : How To Use The Quadratic Function

Example Question #7 : How To Use The Quadratic Function

Example Question #1 : How To Use The Quadratic Function

Example Question #111 : Algebraic Functions

All PSAT Math Resources

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