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SAT Math : Algebra

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Example Questions

Example Question #13 : How To Find F(X)

If f(x) = 4x + 2(x+1) , then f(8x) =?

Possible Answers:

48x + 2

48x + 8

32x + 1

16x + 4

Correct answer:

48x + 2

Explanation:

f(x) = 4x + 2(x+1) ,

f(8x) = 4(8x) + 2(8x+1) = 32x + 16x + 2 = 48x + 2

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Example Question #161 : How To Find F(X)

Define two functions as follows:

f(x) = 4x- 17

$g(x) = \frac{1}{2}x+ 13$

Evaluate $(g \circ f ) (10)$ .

Possible Answers:

$-12 \frac{1}{2}$

$24 \frac{1}{2}$

$2 \frac{3}{4}$

$\textup{None of the other choices gives the correct response}$

Correct answer:

$24 \frac{1}{2}$

Explanation:

By definition, $(g \circ f ) (10) = g [f(10)]$ .

First, evaluate f(10) by setting x = 10 in the definition of f(x) :

f(x) = 4x- 17

$f(10) = 4 \cdot 10 - 17= 40 - 17 = 23$

$(g \circ f ) (10) = g [f(10)] = g(23)$ , so evaluate g(23) similarly:

$g(x) = \frac{1}{2}x+ 13$

$g(23 ) = \frac{1}{2} \cdot 23 + 13 = 11 \frac{1}{2} + 13 = 24 \frac{1}{2}$

$(g \circ f ) (10) = 24 \frac{1}{2}$

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Example Question #162 : How To Find F(X)

Define $f(x) = \sqrt{2x+ 3}$ , restricting the domain of the function to the interval [11, 39 ] .

Give the range of the function.

Possible Answers:

None of these

[-9, 9 ]

[5, 9 ]

$[-9, -5 ] \cup [5, 9 ]$

$(-\infty, \infty )$

Correct answer:

[5, 9 ]

Explanation:

If $11 \le x \le 39$ , then, by way of the properties of inequality, we can multiply all expressions by 2:

$2 \cdot 11 \le 2 \cdot x \le 2 \cdot 39$

$22 \le 2 x \le 78$

then add 3 to all expressions:

$22+ 3 \le 2 x+ 3 \le 78 + 3$

$25 \le 2 x+ 3 \le 81$

Taking the square root of all expressions, we get

$\sqrt{25 }\le \sqrt{2 x+ 3} \le \sqrt{81}$

$5 \le \sqrt{2 x+ 3} \le 9$

$5 \le f(x) \le 9$ .

The correct range is [5, 9 ] .

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Example Question #201 : Algebraic Functions

Define two functions as follows:

f(x) = 4x- 17

$g(x) = \frac{1}{2}x+ 13$

Evaluate:

$(f \circ g ) (10)$

Possible Answers:

$24 \frac{1}{2}$

$-12 \frac{1}{2}$

$2 \frac{3}{4}$

None of these

Correct answer:

Explanation:

By definition, $(f \circ g ) (10) = f [g(10)]$ .

First, evaluate g(10) by setting x = 10 in the definition of g(x) :

$g(x) = \frac{1}{2}x+ 13$

$g(10) = \frac{1}{2} \cdot 10 + 13 = 5 + 13 = 18$

$(f \circ g ) (10) = f [g(10)] = f(18)$ , so evaluate f(18) similarly:

f(x) = 4x- 17

$f(18) = 4 \cdot 18 - 17= 72 - 17 = 55$

$(f \circ g ) (10) = 55$

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Example Question #33 : Algebraic Functions

The function is defined as f(x) = 3x^2 + 6x + 27 . What is f(-1) ?

Possible Answers:

-36

Correct answer:

Explanation:

Substitute -1 for in the given function.

$f(-1)=3(-1)^{2}+6(-1)+27$

f(-1)=3(1)-6+27

f(-1)=3-6+27

f(-1)=24

If you didn’t remember the negative sign, you will have calculated 36. If you remembered the negative sign at the very last step, you will have calculated -36; however, if you did not remember that (-1)^2 is 1, then you will have calculated 18.

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Example Question #34 : Algebraic Functions

If the function h(x) is created by shifting f(x) up four units and then reflecting it across the x-axis, which of the following represents h(x) in terms of f(x) ?

Possible Answers:

f(x)+4

f(x+4)

-f(x)-4

f(x-4)

-f(x+4)

Correct answer:

-f(x)-4

Explanation:

We can take each of the listed transformations of f(x) one at a time. If f(x) is to be shifted up by four units, increase every value of f(x) by 4.

$f(x)_{\textup{shifted up by 4}}=f(x)+4$

Next, take this equation and reflect it across the x-axis. If we reflect a function across the x-axis, then all of its values will be multiplied by negative one. So, h(x) can be written in the following way:

h(x)=-(f(x)+4)

Lastly, distribute the negative sign to arrive at the final answer.

h(x)=-f(x)-4

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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 = -4, x = 2

x = -3, x = 4

x = -5, x = 1

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:

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

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

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

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

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

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 #1 : How To Use The Quadratic Function

If $x^{2}-12x+36=9$ , which two values of are correct?

Possible Answers:

x=12, x=3

x=5, x=7

x=3, x=9

x=-6, x=7

x=4, x=8

Correct answer:

x=3, x=9

Explanation:

First, we set the quadratic function equal to :

$x^{2}-12x+36=9$

$x^{2}-12x+27=0$

Reduce the function to its two component factors:

(x-3)(x-9)=0

Therefore, since either x-3=0 or x-9=0 ,

x=3, x=9

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Example Question #2 : How To Use The Quadratic Function

If $x^{2}+5x-3=-9$ , which pair of values for are correct?

Possible Answers:

x=3, x=-2

x=5, x=-2

x=-3, x=2

x=3, x=2

x=-3, x=-2

Correct answer:

x=-3, x=-2

Explanation:

First, set the quadratic function equal to :

$x^{2}+5x-3=-9$

$x^{2}+5x+6=0$

Then, reduce the function to its two factors:

(x+3)(x+2)=0

Since one of the factors on the left hand side of the equation must equal in order for the above equation to be true,

x+3=0 or x+2=0

Solving for both, we get x=3, x=2 .

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SAT Math : Algebra

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #13 : How To Find F(X)

Example Question #161 : How To Find F(X)

Example Question #162 : How To Find F(X)

Example Question #201 : Algebraic Functions

Example Question #33 : Algebraic Functions

Example Question #34 : Algebraic Functions

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

All SAT Math Resources

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