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Flashcards: Graphical Representation of Functions

Find the zeros of the following function.

$\displaystyle x^2+5x+6$

x=-2 $\displaystyle x=-2$

x=2,3 $\displaystyle x=2,3$

x=-2,-3 $\displaystyle x=-2,-3$

x=-2,3 $\displaystyle x=-2,3$

x=2,-3 $\displaystyle x=2,-3$

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Flashcards: Graphical Representation of Functions

Find the zeros of the following function.

$\displaystyle x^2+5x+6$

x=-2 $\displaystyle x=-2$

x=2,3 $\displaystyle x=2,3$

x=-2,-3 $\displaystyle x=-2,-3$

x=-2,3 $\displaystyle x=-2,3$

x=2,-3 $\displaystyle x=2,-3$

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Flashcards: Graphical Representation of Functions

Find the zeros of the following function.

$\displaystyle x^2+5x+6$

Your answer:

x=-2 $\displaystyle x=-2$

Your answer:

x=2,3 $\displaystyle x=2,3$

Your answer:

x=-2,3 $\displaystyle x=-2,3$

Your answer:

x=2,-3 $\displaystyle x=2,-3$

Correct answer:

x=-2,-3 $\displaystyle x=-2,-3$

Explanation:

To find the zeros of the function use factoring.

$\displaystyle x^2+5x+6$

Set up the expression in factored form, leaving blanks for the numbers that are not yet known.

$(x+\underline{\hspace{.5cm}})(x+\underline{\hspace{.5cm}})$ $\displaystyle (x+\underline{\hspace{.5cm}})(x+\underline{\hspace{.5cm}})$

At this point, you need to find two numbers - one for each blank. By looking at the original expression, a few clues can be gathered that will help find the two numbers. The product of these two numbers will be equal to the last term of the original expression (6, or c in the standard quadratic formula), and their sum will be equal to the coefficient of the second term of the original expression (5, or b in the standard quadratic formula). Because their product is positive (6) and the sum is positive, that must mean that they both have positive signs.

Now, at this point, test a few different possibilities using the clues gathered from the original expression. In the end, it's found that the only numbers that work are 2 and 3, as the product of 2 and 3 is 6, and sum of 2 and 3 is 5. So, this results in the expression's factored form looking like...

$\displaystyle (x+2)(x+3)$

From here, set each binomial equal to zero and solve for $\displaystyle x$.

$\\x+2=0 \\x+2-2=-2 \\x=-2$ $\displaystyle \\x+2=0 \\x+2-2=-2 \\x=-2$

and

$\\x+3=0 \\x+3-3=-3 \\x=-3$ $\displaystyle \\x+3=0 \\x+3-3=-3 \\x=-3$

To verify the zeros, graph the original function and identify where the graph touches or crosses the x-axis.

Therefore the zeros of the function are,

x=-2,-3 $\displaystyle x=-2,-3$

Flashcards: Graphical Representation of Functions

Find all possible zeros for the following function.

x^2-9 $\displaystyle x^2-9$

x=0 $\displaystyle x=0$

x=1 $\displaystyle x=1$

x=-3, 3 $\displaystyle x=-3, 3$

x=-3 $\displaystyle x=-3$

x=3 $\displaystyle x=3$

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Flashcards: Graphical Representation of Functions

Find all possible zeros for the following function.

x^2-9 $\displaystyle x^2-9$

Your answer:

x=0 $\displaystyle x=0$

Your answer:

x=1 $\displaystyle x=1$

Your answer:

x=-3 $\displaystyle x=-3$

Your answer:

x=3 $\displaystyle x=3$

Correct answer:

x=-3, 3 $\displaystyle x=-3, 3$

Explanation:

To find the zeros of the function, use factoring.

x^2-9 $\displaystyle x^2-9$

Set up the expression in factored form, leaving blanks for the numbers that are not yet known.

$(x+\underline{\hspace{.5cm}})(x+\underline{\hspace{.5cm}})$ $\displaystyle (x+\underline{\hspace{.5cm}})(x+\underline{\hspace{.5cm}})$

At this point, you need to find two numbers - one for each blank. By looking at the original expression, a few clues can be gathered that will help find the two numbers. The product of these two numbers will be equal to the last term of the original expression (-9, or c in the standard quadratic formula), and their sum will be equal to the coefficient of the second term of the original expression (0, or b in the standard quadratic formula). Because their product is negative (-9) and the sum is zero, that must mean that they have different signs but the same absolute value.

Now, at this point, test a few different possibilities using the clues gathered from the original expression. In the end, it's found that the only numbers that work are 3 and -3, as the product of 3 and -3 is -9, and sum of 3 and -3 is 0. So, this results in the expression's factored form looking like...

$\displaystyle (x-3)(x+3)$

This is known as a difference of squares.

From here, set each binomial equal to zero and solve for $\displaystyle x$.

$\\x-3=0\\x-3+3=+3\\x=3$ $\displaystyle \\x-3=0\\x-3+3=+3\\x=3$

and

$\\x+3=0\\x+3-3=-3\\x=-3$ $\displaystyle \\x+3=0\\x+3-3=-3\\x=-3$

To verify the zeros, graph the original function and identify where the graph touches or crosses the x-axis.

Q10

Therefore the zeros of the function are,

x=-3,3 $\displaystyle x=-3,3$

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