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SAT Math : SAT Mathematics

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Example Questions

Example Question #51 : 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 #2911 : Sat Mathematics

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 #2912 : Sat Mathematics

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

Possible Answers:

x = -4, x = -2

x = -4, x = 2

x = -5, x = 1

x = -3, x = 4

x = 8, x = 0

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:

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

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

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

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

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

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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Example Question #2 : How To Use The Quadratic Function

If $x^{2}+x-50=-8$ , which two values of are correct?

Possible Answers:

x=-25, x=2

x=6, x=-7

x=-6, x=7

x=25, x=-2

x=5, x=-3

Correct answer:

x=6, x=-7

Explanation:

First, set the quadratic function equal to :

$x^{2}+x-50=-8$

$x^{2}+x-42=0$

Then, separate the function into its two component factors:

(x-6)(x+7)=0

It follows from this equation that either x-6=0 or x+7=0 ,

x=6, x=-7

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Example Question #2 : How To Use The Quadratic Function

The length of a rectangular piece of land is two feet more than three times its width. If the area of the land is $125\:ft^2$ , what is the width of that piece of land?

Possible Answers:

$\frac{1-2\sqrt{94}}{3}\:ft^2$

$\frac{-1-2\sqrt{94}}{3}\:ft^2$

$\frac{2-4\sqrt{94}}{6}\:ft^2$

$\frac{-1+4\sqrt{94}}{3}\:ft^2$

$\frac{-1+2\sqrt{94}}{3}\:ft^2$

Correct answer:

$\frac{-1+2\sqrt{94}}{3}\:ft^2$

Explanation:

The area of a rectangle is the product of its length by its width, which we know to be equal to $125\:ft^2$ in our problem. We also know that the length is equal to 2+3w , where represents the width of the land. Therefore, we can write the following equation:

w(2+3w)=125

Distributing the outside the parentheses, we get:

$3w^{2}+2w=125$

Subtracting 125 from each side of the equation, we get:

$3w^{2}+2w-125=0$

We get a quadratic equation, and since there is no factor of and that adds up to , we use the quadratic formula to solve this equation.

a=3 b=2 c=-125

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

We can first calculate the discriminant (i.e. the part under the square root)

$b^{2}-4ac= 2^{2}-4\times3 \times(-125)$

$4-4\times3\times(-125)$

$4-12\times(-125)$

4-(-1500)=1504

We replace that value in the quadratic formula, solving both the positive version of the formula (on the left) and the negative version of the formula (on the right):

$w=\frac{-2+\sqrt{1504}}{6}$ $w=\frac{-2+\sqrt{1504}}{6}$

Breaking down the square root:

$w=\frac{-2+\sqrt{2\times2\times2\times2\times2\times 47}}{6}$ $w=\frac{-2-\sqrt{2\times2\times2\times2\times2\times 47}}{6}$

We can pull two of the twos out of the square root and place a outside of it:

$w=\frac{-2+4\sqrt{2\times 47}}{6}$ $w=\frac{-2-4\sqrt{2\times 47}}{6}$

We can then multiply the and the :

$w=\frac{-2+4\sqrt{94}}{6}$ $w=\frac{-2-4\sqrt{94}}{6}$

At this point, we can reduce the equations, since each of the component parts of their right sides has a factor of :

$w=\frac{-1+2\sqrt{94}}{3}$ $w=\frac{-1+2\sqrt{94}}{3}$

Since width is a positive value, the answer is:

$w=\frac{-1+2\sqrt{94}}{3}$

$w=\frac{-1+2\times9.695}{3}$

$w=\frac{-1+19.39}{3}$

$w=\frac{18.39}{3}=6.13$

The width of the piece of land is approximately $6.13\:ft^2$ .

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Example Question #2913 : Sat Mathematics

f(x)=2(x-3)^2+5

What is the focus of the above quadratic equation?

Possible Answers:

$\left(3, 5\right)$

$\left(3, \frac{41}{8}\right)$

$\left(-3, \frac{41}{8}\right)$

$\left(3, \frac{1}{8}\right)$

$\left(-3, -\frac{41}{8}\right)$

Correct answer:

$\left(3, \frac{41}{8}\right)$

Explanation:

The focus is solved by using the following formula (x,y+p) , where x,y are the coordinates of the vertex, and is the distance from the vertex. To solve for , we use

$a=\frac{1}{4p}$ , where is the coefficient in front of the x^2 term and is the focus.

Since a=2 , we can substitute in to get,

$2=\frac{1}{4p}$

$2p=\frac{1}{4}$

$p=\frac{1}{4}\cdot\frac{1}{2}=\frac{1}{8}$

Now we need to find the vertex of the equation.

Since our equation is in vertex form, we can deduce that the vertex is at (3,5)

So the focus is at $\left( 3,5+\frac{1}{8}\right)=\left( 3,\frac{41}{8}\right)$

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Example Question #2914 : Sat Mathematics

Find all the solutions of where x^3+1 crosses the line x+1 .

Possible Answers:

x=0

x=1

x=-1

No Real Solutions

x=0

x=-1

x=0

x=1

x=-1

Correct answer:

x=0

x=1

x=-1

Explanation:

In order to find all the solutions, we need to set the equations equal to each other.

x^3+1=x+1

Now subtract and from each side.

x^3-x+1-1=0

x^3-x=0

Factor the left hand side to get

x(x^2-1)=0

Factor the quadratic function inside the parenthesis to get

x(x-1)(x+1)=0

The solutions to this equation are

x=0

$x-1=0\rightarrow x=1$

$x+1=0\rightarrow x=-1$

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SAT Math : SAT Mathematics

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #51 : Algebraic Functions

Example Question #2911 : Sat Mathematics

Example Question #2912 : Sat Mathematics

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 #2 : How To Use The Quadratic Function

Example Question #2 : How To Use The Quadratic Function

Example Question #2913 : Sat Mathematics

Example Question #2914 : Sat Mathematics

All SAT Math Resources

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