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PSAT Math : Algebra

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Example Questions

Example Question #5 : How To Find The Domain Of A Function

What is the domain of the function?

$f(x) = 2x^{2}-11x+5$

Possible Answers:

All real numbers

$\frac{1}{2} < x < 5$

$x<\frac{1}{2}\ \text{and}\ x>5$

All positive integers

$x=\frac{1}{2},\ 5$

Correct answer:

All real numbers

Explanation:

The domain of f(x) is all the values of x for which f(x) is defined.

f(x) has no square roots or denominators, so it will always be defined; there are no restrictions on x because any and all values will lead to a real result.

Therefore, the domain is the set of all real numbers.

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Example Question #7 : How To Find The Domain Of A Function

Which of the following represents the domain of f(x) where:

$f(x)=(x-2)^{\frac{4}{5}}+5$

Possible Answers:

x<2

x>0

x<0

x>3

is all real numbers

Correct answer:

is all real numbers

Explanation:

Using our properties of exponents, we could rewrite f(x) as $\sqrt[5]{(x-2)^{4}}+3$

This means that when we input , we first subtract 2, then take this to the fourth power, then take the fifth root, and then add three. We want to look at these steps individually and see whether there are any values that wouldn’t work at each step. In other words, we want to know which values we can put into our function at each step without encountering any problems.

The first step is to subtract 2 from . The second step is to take that result and raise it to the fourth power. We can subtract two from any number, and we can take any number to the fourth power, which means that these steps don't put any restrictions on .

Then we must take the fifth root of a value. The trick to this problem is recognizing that we can take the fifth root of any number, positive or negative, because the function $x^{\frac{1}{5}}$ is defined for any value of ; thus the fact that f(x) has a fifth root in it doesn't put any restrictions on , because we can add three to any number; therefore, the domain for f(x) is all real values of .

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Example Question #8 : How To Find The Domain Of A Function

What is the domain of the given function?

$f(x)=\frac{x+6}{({x+3})^2}$

Possible Answers:

x ≠ –3

All real numbers

x ≠ 0

x ≠ 3

x ≠ 3, –3

Correct answer:

x ≠ –3

Explanation:

The domain of the function is all real numbers except x = –3. When x = –3, f(–3) is undefined.

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Example Question #1 : How To Find The Domain Of A Function

Find the domain of the given function:

$f(x)=\frac{(x+3)^2}{(x-3)(x)}$

Possible Answers:

All real numbers x such that x ≠ 3, 0

All real numbers x such that x ≠ 1, 0

All real numbers x such that x ≠ 0

All real numbers

All real numbers x such that x ≠ 3

Correct answer:

All real numbers x such that x ≠ 3, 0

Explanation:

When x = 0 or x = 3, the function is undefined due to its denominator.

Thus the domain is all real numbers x, such that x is not equal to 0 or 3.

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Example Question #1 : Algebraic Functions

Find the domain of the function:

$f(x)=\frac{x^2-1}{x-1}$

Possible Answers:

All real numbers except for –2

All real numbers except for 1

All real numbers

Correct answer:

All real numbers except for 1

Explanation:

If a value of x makes the denominator of a equation zero, that value is not part of the domain. This is true, even here where the denominator can be "cancelled" by factoring the numerator into

$\left ( x+1\right )\left ( x-1\right )$

and then cancelling the $\left ( x-1\right )$ from the numerator and the denominator.

This new expression, f(x)=x+1 is the equation of the function, but it will have a hole at the point where the denominator originally would have been zero. Thus, this graph will look like the line y=x+1 with a hole where x-1=0 , which is

x=1 .

Thus the domain of the function is all values such that $x\neq1$

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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 = -5, x = 1

x = -4, x = 2

x = 8, x = 0

x = -4, x = -2

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 #2 : How To Use The Quadratic Function

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

Possible Answers:

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

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

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

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

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

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 #2 : 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:

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

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

None of the other answers

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

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

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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PSAT Math : Algebra

Study concepts, example questions & explanations for PSAT Math

All PSAT Math Resources

Example Questions

Example Question #5 : How To Find The Domain Of A Function

Example Question #7 : How To Find The Domain Of A Function

Example Question #8 : How To Find The Domain Of A Function

Example Question #1 : How To Find The Domain Of A Function

Example Question #1 : Algebraic Functions

Example Question #1 : How To Use The Quadratic Function

Example Question #2 : 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

All PSAT Math Resources

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