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Find the highest point of the parabola

Great, let’s go with more practical cases, a tunnel. Because now you first build the tunnel and then determine its highest point, has not it happened to you? Let us begin!

A tunnel has a parabolic shape such that its equation can be written in the form x^{2} - 8x + 8y - 32 = 0 , determine the highest point.

parabola-tunnel
The parabolic tunnel.

Well, to determine the highest point, we’re going to take the equation that they gave us and we’re going to leave the x‘s on one side of the equation and the y‘s on the other side of the equation:

x^{2}-8x = -8y+32

In order to continue with the resolution to find the canonical equation of the parabola, you have to know a skill to complete the binomial squared without having to test numbers.

Okay, we know very well that the binomial to the perfect square says that a binomial squared equals the square of the first term plus the double product of the first term by the second plus the square of the second term, that is:

(a+b)^{2} = a^{2} + 2ab + b^{2}

Then, to know what number we are going to add to complete the binomial, we will see that -8x is the part of the result of the binomial, that is, the double product of the first term by the second, in mathematical terms it is 2ab.

Well, now what we have to do is equalize that 2ab with -8x:

2ab = -8x

What we need is to find the second term, that is, b because we already know what is a, a is equal to x, so we substitute a with x:

2xb = -8x

Now we will clear to obtain the value of the second term:

b = \cfrac{-8x}{2x}

We will cancel the x‘s:

b = \cfrac{-8}{2}

And -8 divided by 2 equals -4:

b = -4

Now that we have the value of the second term, we just have to complete it in the equation that they gave us in the parabola, that means that x^{2}-8x we are going to add a 16 so that the binomial is complete, in the same way we have to add a 16 on the other side of the equation so that it is balanced:

x^{2}-8x+16=-8y+32+16

We will sum the 32 with the 16:

x^{2}-8x+16=-8y+48

Now yes, the algebraic identity of x^{2}-8x+16 is:

(x-4)^{2}

We will write it in the equation and we will factor a -8 on the side of the equation where the y are:

(x-4)^{2}=-8(y-6)

Which gives us as a result that the highest point of the parabola is equal to the vertex because the parabola is vertical that opens downwards:

V(4,6)

In addition, the parabola obtained is logical because what we are supposed to be working with is a tunnel and a tunnel of today is not some parabola that opens up or that is horizontal and that opens to the right or to the left. Obviously the parabola obtained has to open down, can you imagine a parabolic tunnel that opens to the right?

Thank you for being at this moment with us :)

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