1. ## Question for Maths

Hey guys, could you help me out with this?

--------------------------------------...

Given f: (-3 , 6] where $y = ( x^2 - 3) ( x - 4 )$

a) Find the coordinates of the x and y intercepts.

b) Find the coordinates of the local minium

--------------------------------------...

Thanks

2. Originally Posted by mibamars
Hey guys, could you help me out with this?

--------------------------------------...

Given f: (-3 , 6] where $y = ( x^2 - 3) ( x - 4 )$

a) Find the coordinates of the x and y intercepts.

b) Find the coordinates of the local minium

--------------------------------------...

Thanks
a) As $x^{2}-3 \mbox{ and } x-4$ are being multiplied by eachother, if either of them equals zero, the equation equals zero, and thus y equals zero, and you have an x intercept.

so set each equation to zero
$x^{2}-3=0$
$x^{2}=3$
$x=\pm\sqrt{3}$

and
$x-4=0$
$x=4$

So you have x-intercepts at $(-\sqrt{3},0), (\sqrt{3}, 0), (4, 0)$

And you determine your y-intercept by setting x to zero, so
$y=(x^{2}-3)(x-4)$
$y=(0^{2}-3)(0-4)$
$y=(-3)(-4)$
$y=12$

So our y-intercept is $(0,12)$

B)Local maximums and minimums occur at critical points, so find the critical points (the slope goes to zero or infinity) This can be determined by finding the equation of the slope (the first derivative)
$y=(x^{2}-3)(x-4)$
$y=x^{3}-4x^{2}-3x+12$
$y\prime=3x^{2}-8x-3$
$y\prime=(3x+1)(x-3)$

So there are no points where this goes to infinity, but it does have points where it goes to zero, using same method from step a
$3x+1=0$
$x=\frac{-1}{3}$

and
$x-3=0$
$x=3$

Now we just need to determine whether our x values are maximums or minimums. There are really 3 ways to do this, you can check the y values around it on your graph of the function, you can check the sign of points around this point on your first derivative, and you can check your sign on your second derivative. We'll do the second derivative, because it is less susceptible to error that comes from choosing adjacent points over the actual point.

So
$y\prime=3x^{2}-8x-3$
$y\prime\prime=6x-8$

And plug our values in
$y\prime\prime(\frac{-1}{3})=6\frac{-1}{3}-8$
$y\prime\prime(\frac{-1}{3})=-2-8$
$y\prime\prime(\frac{-1}{3})=-10$

This point has a negative slope, this means it is concave down, so the slope goes down away from our point on both sides, this means that our point is higher than everything around it, so it is a local maximum, not what we need, so lets check our other value.
$y\prime\prime(3)=6(3)-8$
$y\prime\prime(3)=18-8$
$y\prime\prime(3)=10$

This is positive, so our function is concaved up here, which means on either side of our point, the slope goes up, so this point is below everything around it, making it a local minimum

So now we just need to plug it into our function and find it's y value.
$y(x)=(x^{2}-3)(x-4)$
$y(3)=((3)^{2}-3)(3-4)$
$y(3)=(9-3)(-1)$
$y(3)=(6)(-1)$
$y(3)=-6$

So our local minimum is at (3,-6)

3. Originally Posted by mibamars
Hey guys, could you help me out with this?

--------------------------------------...

Given f: (-3 , 6] where $y = ( x^2 - 3) ( x - 4 )$

a) Find the coordinates of the x and y intercepts.

b) Find the coordinates of the local minium
Hello,

to a):
y-intercept: Plug in x = 0 and calculate the value of y

x-intercept: Plug in y = 0 and calculate for x. In your case: A product equals zero if one of the factors equal zero. Thus you have to solve for x:
$x^2-3=0~\vee~x-4=0$

to b) With your problem it would be easier to use the expanded form of your function:

$y = x^3-4x^2-3x +12~\implies~y'=3x^2-8x-3$ Plug in y' = 0 and solve for x (-1/3, 3)

Plug in these values into the equation of y.