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Flashcards: Use the zeros to construct a rough graph of a function

The graph of a function is shown below, with labels on the y-axis hidden.

Graph zeroes 2

Determine which of the following functions best fits the graph above.

$\displaystyle f(x)=-2x^2(x+1)(x-2)(x+2)$

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

$\displaystyle f(x)=-2(x+1)(x-2)(x+2)$

$\displaystyle f(x)=(x+1)^3(x-2)^2$

$f(x)=\frac{x+1}{x+2}$ $\displaystyle f(x)=\frac{x+1}{x+2}$

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Flashcards: Use the zeros to construct a rough graph of a function

The graph of a function is shown below, with labels on the y-axis hidden.

Graph zeroes 2

Determine which of the following functions best fits the graph above.

$\displaystyle f(x)=-2x^2(x+1)(x-2)(x+2)$

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

$\displaystyle f(x)=-2(x+1)(x-2)(x+2)$

$\displaystyle f(x)=(x+1)^3(x-2)^2$

$f(x)=\frac{x+1}{x+2}$ $\displaystyle f(x)=\frac{x+1}{x+2}$

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Flashcards: Use the zeros to construct a rough graph of a function

Determine which of the following functions best fits the graph above.

Your answer:

$\displaystyle f(x)=-2x^2(x+1)(x-2)(x+2)$

Your answer:

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

Your answer:

$\displaystyle f(x)=(x+1)^3(x-2)^2$

Your answer:

$f(x)=\frac{x+1}{x+2}$ $\displaystyle f(x)=\frac{x+1}{x+2}$

Correct answer:

$\displaystyle f(x)=-2(x+1)(x-2)(x+2)$

Explanation:

Use the zeroes of the graph to determine the matching function. Zeroes are values of x where f(x)=0 $\displaystyle f(x)=0$. In other words, they are points on the graph where the curve touches zero.

Visually, you can see that the curve crosses the x-axis when x=-2 $\displaystyle x=-2$, x=-1 $\displaystyle x=-1$, and x=2 $\displaystyle x=2$. Therefore, you need to look for a function that will equal zero at these x values.

A function with a factor of (x+2) $\displaystyle (x+2)$ will equal zero when x=-2 $\displaystyle x=-2$, because the factor of (x+2) $\displaystyle (x+2)$ will equal zero. The matching factors for the other two zeroes, x=-1 $\displaystyle x=-1$ and x=2 $\displaystyle x=2$, are (x+1) $\displaystyle (x+1)$ and (x-2) $\displaystyle (x-2)$, respectively.

The answer choice $\displaystyle f(x)=-2x^2(x+1)(x-2)(x+2)$ has all of these factors, but it is not the answer because it has an additional zero that would be visible on the graph. Notice it has a factor of x^2 $\displaystyle x^2$, which results in a zero at x=0 $\displaystyle x=0$. This additional zero that isn't present in the graph indicates that this cannot be matching function.

$\displaystyle f(x)=-2(x+1)(x-2)(x+2)$ is the answer because it has all of the required factors and, as a result, the required zeroes, while not having additional zeroes. Notice that the constant coefficient of negative 2 does not affect where the zeroes are.

Flashcards: Use the zeros to construct a rough graph of a function

Graph zeroes

Which of the functions below best matches the graphed function?

$\displaystyle f(x)= x(3x-5)(x-2)$

$\displaystyle f(x)= x(3x-5)(x+2)$

$\displaystyle f(x)= (3x+5)(x+2)$

$\displaystyle f(x)= x(x+2)^2$

$\displaystyle f(x)= x(3x+5)(x-2)$

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Flashcards: Use the zeros to construct a rough graph of a function

Which of the functions below best matches the graphed function?

Your answer:

$\displaystyle f(x)= x(3x-5)(x-2)$

Your answer:

$\displaystyle f(x)= (3x+5)(x+2)$

Your answer:

$\displaystyle f(x)= x(x+2)^2$

Your answer:

$\displaystyle f(x)= x(3x+5)(x-2)$

Correct answer:

$\displaystyle f(x)= x(3x-5)(x+2)$

Explanation:

First, look at the zeroes of the graph. Zeroes are where the function touches the x-axis (i.e. values of $\displaystyle x$ where $\displaystyle f(x)=0$).

The graph shows the function touching the x-axis when $\displaystyle x=-2$, $\displaystyle x=0$, and at a value in between 1.5 and 2.

Notice all of the possible answers are already factored. Therefore, look for one with a factor of $\displaystyle x$ (which will make $\displaystyle f(x)=0$ when $\displaystyle x=0$), a factor of $\displaystyle (x+2)$ to make $\displaystyle f(x)=0$ when $\displaystyle x=-2$, and a factor which will make $\displaystyle f(x)=0$ when $\displaystyle x$ is at a value between 1.5 and 2.

$\displaystyle f(x)= x(3x-5)(x+2)$

This function fills the criteria; it has an $\displaystyle x$ and an $\displaystyle (x+2)$ factor. Additionally, the third factor, $\displaystyle (3x-5)$, will result in $\displaystyle f(x)=0$ when $x=\frac{5}{3}$ $\displaystyle x=\frac{5}{3}$, which fits the image. It also does not have any extra zeroes that would contradict the graph.

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