WEBVTT
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Which of the following functions corresponds to a graph in the π₯π¦-plane with π₯-intercepts at negative three and five and a π¦-intercept of 30? a) π of π₯ is equal to six π₯ plus two multiplied by π₯ minus five. b) π of π₯ is equal to two π₯ plus six multiplied by π₯ plus five. c) π of π₯ is equal to negative two π₯ minus six multiplied by π₯ minus five.
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Or d) π of π₯ is equal to π₯ plus three multiplied by π₯ minus five.
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So the first thing we want to take a look at is the fact that we have π₯-intercepts at negative three and five.
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That means that our function will have zeros at negative three and five.
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So to check the zeros of our functions, so our possible four answers, what we can do is set them equal to zero.
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So first of all, for a, we have six π₯ plus two multiplied by π₯ minus five is equal to zero.
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Now to calculate what the zeros are or the π₯-values, what we need to do is make one of our parentheses equal to zero, because at least one of them needs to be equal to zero to have a result of zero, because zero multiplied by anything gives us zero.
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So first of all, weβre gonna set the left-hand parenthesis to zero.
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So weβve got six π₯ plus two equals zero.
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So then if I subtract two from each side of the equation, Iβm gonna get six π₯ is equal to negative two.
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So then what I can do is divide each side of the equation by six to find out what π₯ would be.
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So we can find out that one of our π₯-values, so one of our zeros, is gonna be equal to negative two over six for this function.
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Well, negative two over six is not negative three or five because actually if we simplified it, itβd be equal to negative a third.
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So therefore, we can say that function a is definitely not the correct function.
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So now what we can do is move on to function b.
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So again, weβre gonna set it equal to zero.
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So we have two π₯ plus six multiplied by π₯ plus five is equal to zero.
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So then if we have two π₯ plus six equals zero, what we need to do first of all is subtract six from each side of the equation to solve this.
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So we have two π₯ is equal to negative six.
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And then we can divide by two to find out our value for π₯.
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And when we do that, we get π₯ is equal to negative three.
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So great, this is one of the π₯-intercepts that weβre looking for.
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So function b is possible.
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So what we can do is look at the second parenthesis.
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So if we set this equal to zero, we get π₯ plus five is equal to zero.
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So then we can subtract five from each side of the equation.
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And we get π₯ is equal to negative five.
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So this is not the correct function either because our π₯-intercepts are negative three and five, not negative three and negative five.
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So now we can move on to c.
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And for c, weβve got negative two π₯ minus six multiplied by π₯ minus five is equal to zero.
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So now if we look at the left-hand side, to make this equal to zero, weβve got negative two π₯ minus six is equal to zero.
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So then if we add six to each side of the equation, we get negative two π₯ equals six.
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So then what we can do is divide each side of the equation by negative two to find π₯.
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And when we do that, we get π₯ is equal to negative three.
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And thatβs because six divided by negative two is negative three, because if we have a positive divided by a negative, itβs a negative.
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So great, this actually fits with one of our π₯-intercepts.
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So letβs check out the right-hand parenthesis.
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So we have π₯ minus five is equal to zero.
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So if we add five to each side of the equation, we get π₯ is equal to five.
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So great, this matches the other π₯-intercept.
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So therefore, function c could be the correct function.
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So finally, we move on to function d.
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Now for function d, weβve got π₯ plus three multiplied by π₯ minus five is equal to zero.
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So weβll start with the left-hand side again.
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So we get π₯ plus three is equal to zero.
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So if I subtract three from each side of the equation, I get π₯ is equal to negative three.
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So great, this is one of our π₯-intercepts.
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And then on the other side, Iβve got π₯ minus five is equal to zero.
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And then Iβll add five to each side of the equation.
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And this would give us π₯ is equal to five.
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So this is also one of the π₯-intercepts that weβre looking for.
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So we now know that the possible functions that could be the function that weβre looking for are c or d.
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So now what we need to do is decide which one it is.
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Is it c or is it d?
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And to do this, what weβre going to do is to use the fact that we know that the π¦-intercept is 30.
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So what this means is that we know that our function is going to cross the π¦-intercept at a π¦-value of 30.
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So therefore, itβs gonna have a coordinate of zero, 30, because the π₯-coordinate would be zero at the π¦-axis.
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So therefore, to actually see whether we have the correct function, what I can do is substitute in π₯ is equal to zero to see if the function will be equal to 30 when this is the case.
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So for c), what weβre gonna have is negative two multiplied by zero minus six.
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Thatβs because we substituted our π₯ for zero.
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And then this is multiplied by zero minus five.
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So this is gonna leave us with negative six multiplied by negative five.
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And thatβs because negative two multiplied by zero is just zero.
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So zero minus six is just negative six.
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So as weβve said, weβve got negative six multiplied by negative five.
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Well, we know that a negative multiplied by a negative is equal to a positive.
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And we know that six multiplied by five is 30.
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So we can say that the value of the function of c is going to be 30 when π₯ is equal to zero.
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So this is what we were looking for, because we wanted a π¦-intercept of 30.
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So we can say that we think that, yes, c is going to be the correct function.
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But we can double-check this by checking d.
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For d, weβre gonna have zero plus three multiplied by zero minus five.
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So this is gonna give us three multiplied by negative five, which will give us a π¦-value or the function value of negative 15.
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So this is incorrect because weβre looking for a π¦-intercept of 30.
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So therefore, we can say that the correct function that corresponds to a graph in the π₯π¦-plane with π₯-intercepts at negative three and five and a π¦-intercept of 30 is function c π of π₯ is equal to negative two π₯ minus six multiplied by π₯ minus five.