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SAT Math : How to use the quadratic function

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Example Questions

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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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 #3 : 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:

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

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

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

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

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

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 #2 : How To Use The Quadratic Function

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

What is the focus of the above quadratic equation?

Possible Answers:

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

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

$\left(3, 5\right)$

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

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

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 #5 : How To Use The Quadratic Function

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

Possible Answers:

x=1

x=-1

No Real Solutions

x=0

x=1

x=-1

x=0

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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Example Question #3 : How To Use The Quadratic Function

How many x-intercepts does the following equation have?

f(x)=x^3-x

Possible Answers:

Correct answer:

Explanation:

To figure out how many x-intercepts there are, we need to set out equation equal to zero.

f(x)=x^3-x

0=x^3-x

Factor an x out

0=x(x^2-1)

Factor inside of the parenthesis.

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

We can see there are 3 x-intercepts at x=0, x=1, x=-1 .

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Example Question #7 : How To Use The Quadratic Function

To use a computer at a computer lab, Tonya must pay $0.50 per hour. She spends hours at the computer lab, and pays a total amount of , where C=1.2+0.5h . If this equation is graphed, what does the y-intercept represent?

Possible Answers:

The total amount that Tonya paid.

The number of hours Tonya spent at the computer lab.

The initial cost of using a computer.

The amount per hour that it costs to use a computer.

The amount per day that it costs to use a computer.

Correct answer:

The initial cost of using a computer.

Explanation:

If the equation

C=1.2+.5h

were graphed, the result would be a line with a slope of 0.5 and a y-intercept of 1.2. 0.5 represents the amount Tonya pays per hour to work at a computer, because we know that the rate is $0.50, and C represents the total amount she pays. Because we must add 1.2 (or $1.20) to the total amount of hours multiplied by the hourly rate in order to get the total amount Tonya paid, we can assume there is another charge for renting a computer, in addition to the hourly rate, and that that charge is $1.20. Therefore, the y-intercept of the graph of the equation represents the initial cost of using a computer.

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SAT Math : How to use the quadratic function

Study concepts, example questions & explanations for SAT Math

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Example Questions

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

Example Question #2 : How To Use The Quadratic Function

Example Question #3 : How To Use The Quadratic Function

Example Question #2 : How To Use The Quadratic Function

Example Question #5 : How To Use The Quadratic Function

Example Question #3 : How To Use The Quadratic Function

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