### Video Transcript

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. Or d) π of π₯ is equal to π₯ plus three multiplied by π₯ minus five.

So the first thing we want to take a look at is the fact that we have π₯-intercepts at negative three and five. That means that our function will have zeros at negative three and five. So to check the zeros of our functions, so our possible four answers, what we can do is set them equal to zero.

So first of all, for a, we have six π₯ plus two multiplied by π₯ minus five is equal to zero. 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.

So first of all, weβre gonna set the left-hand parenthesis to zero. So weβve got six π₯ plus two equals zero. So then if I subtract two from each side of the equation, Iβm gonna get six π₯ is equal to negative two. So then what I can do is divide each side of the equation by six to find out what π₯ would be.

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. Well, negative two over six is not negative three or five because actually if we simplified it, itβd be equal to negative a third. So therefore, we can say that function a is definitely not the correct function.

So now what we can do is move on to function b. So again, weβre gonna set it equal to zero. So we have two π₯ plus six multiplied by π₯ plus five is equal to zero. 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. So we have two π₯ is equal to negative six. And then we can divide by two to find out our value for π₯.

And when we do that, we get π₯ is equal to negative three. So great, this is one of the π₯-intercepts that weβre looking for. So function b is possible. So what we can do is look at the second parenthesis. So if we set this equal to zero, we get π₯ plus five is equal to zero. So then we can subtract five from each side of the equation. And we get π₯ is equal to negative five. So this is not the correct function either because our π₯-intercepts are negative three and five, not negative three and negative five.

So now we can move on to c. And for c, weβve got negative two π₯ minus six multiplied by π₯ minus five is equal to zero. 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. So then if we add six to each side of the equation, we get negative two π₯ equals six. So then what we can do is divide each side of the equation by negative two to find π₯.

And when we do that, we get π₯ is equal to negative three. And thatβs because six divided by negative two is negative three, because if we have a positive divided by a negative, itβs a negative. So great, this actually fits with one of our π₯-intercepts. So letβs check out the right-hand parenthesis.

So we have π₯ minus five is equal to zero. So if we add five to each side of the equation, we get π₯ is equal to five. So great, this matches the other π₯-intercept. So therefore, function c could be the correct function.

So finally, we move on to function d. Now for function d, weβve got π₯ plus three multiplied by π₯ minus five is equal to zero. So weβll start with the left-hand side again. So we get π₯ plus three is equal to zero. So if I subtract three from each side of the equation, I get π₯ is equal to negative three. So great, this is one of our π₯-intercepts.

And then on the other side, Iβve got π₯ minus five is equal to zero. And then Iβll add five to each side of the equation. And this would give us π₯ is equal to five. So this is also one of the π₯-intercepts that weβre looking for. So we now know that the possible functions that could be the function that weβre looking for are c or d.

So now what we need to do is decide which one it is. Is it c or is it d? And to do this, what weβre going to do is to use the fact that we know that the π¦-intercept is 30. So what this means is that we know that our function is going to cross the π¦-intercept at a π¦-value of 30. So therefore, itβs gonna have a coordinate of zero, 30, because the π₯-coordinate would be zero at the π¦-axis.

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. So for c), what weβre gonna have is negative two multiplied by zero minus six. Thatβs because we substituted our π₯ for zero. And then this is multiplied by zero minus five. So this is gonna leave us with negative six multiplied by negative five. And thatβs because negative two multiplied by zero is just zero. So zero minus six is just negative six.

So as weβve said, weβve got negative six multiplied by negative five. Well, we know that a negative multiplied by a negative is equal to a positive. And we know that six multiplied by five is 30. So we can say that the value of the function of c is going to be 30 when π₯ is equal to zero. So this is what we were looking for, because we wanted a π¦-intercept of 30.

So we can say that we think that, yes, c is going to be the correct function. But we can double-check this by checking d. For d, weβre gonna have zero plus three multiplied by zero minus five. 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. So this is incorrect because weβre looking for a π¦-intercept of 30.

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.