# Video: US-SAT03S4-Q12-793186831482

In the π₯π¦-plane, the graph of function π has π₯-intercepts at β5, β3, and 3. Which of the following could define π? [A] π(π₯) = (π₯ + 5)(π₯ + 3)(π₯ β 3) [B)] π(π₯) = (π₯ β 5)(π₯ + 3)Β² [C] π(π₯) = (π₯ + 5)(π₯ β 3)Β² [D] π(π₯) = (π₯ β 5)(π₯ β 3)(π₯ + 3)

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### Video Transcript

In the π₯π¦-plane, the graph of the function π has π₯-intercepts at negative five, negative three, and three. Which of the following could define π? A) π of π₯ is equal to π₯ plus five times π₯ plus three times π₯ minus three. B) π of π₯ is equal to π₯ minus five times π₯ plus three squared. C) π of π₯ is equal to π₯ plus five times π₯ minus three squared. Or D) π of π₯ is equal to π₯ minus five times π₯ minus three times π₯ plus three.

So, we are told that the graph of our function has π₯-intercepts at negative five, negative three, and three, which means our graph would cross through the π₯-axis at π₯ equals negative five, π₯ equals negative three, and π₯ equals positive three. And itβs asking which of these equations could define π. Well, notice that these are in factored form. So, theyβre the factors multiplied together that create some sort of polynomial, which would be our graph.

Now, while we might not know what the graph looks like, we do know where it crosses the π₯-axis, which is directly related with the factors. If we took π₯ equals negative five, we could create it into a factor by bringing the negative five over with the π₯. So, if we would do that, we would have π₯ plus five equals zero, which would mean π₯ plus five would be a factor. And notice that π₯ plus five is equal to zero, sometimes theyβre also called zeros of functions.

So, if we know that we need π₯ plus five, we can go ahead and eliminate options B and D cause they have π₯ minus five. Now, letβs look at our other π₯-intercepts. So, we have π₯ equals negative three. So, if we would add three to both sides, π₯ plus three would be a factor. And then, working with the π₯-intercept π₯ equals three, we subtract three, and we have π₯ minus three equals zero. So, π₯ minus three would be a factor.

So, we need factors π₯ plus five, π₯ plus three, and π₯ minus three, which would be option A. Now, letβs make a little side note about these exponents. Notice we have π₯ minus three squared for option C. That just means there are two π₯ minus threes multiplied together, so π₯ minus three times π₯ minus three. And we needed π₯ minus three and an π₯ plus three, so C could be eliminated. So, we could define π as π of π₯ equals π₯ plus five times π₯ plus three times π₯ minus three.