focal point (1 Viewer)

[Caterwaul]

Hi, how do you find the focal point of y=x^2+6x+7
i never really learned that part of maths properly so....

 

[liamkk112]

Hi, how do you find the focal point of y=x^2+6x+7
i never really learned that part of maths properly so....

by focal point im assuming u mean the vertex of the parabola.

the formula for the x-coordinate of the vertex is x = -b/2a, where b is the coefficient of the x^1 term and a is the coefficient of the x^2 term
so in this case b = 6, a = 1, so the vertex's x-coordinate is -6/2 = -3. (reasoning: the vertex is the point halfway between the two x-intercepts or the average of the two x-intercepts , and we know from the quadratic formula that the x-intercepts are of the form x = -b+sqrt(b^2-4ac)/2a and x = -b - sqrt(b^2 -4ac)/2a, so taking the average of these (add them and divide by 2) we get that x = -b/2a is the x-coordinate of the vertex [the square roots will cancel])

now we can plug this into the equation y = x^2 +6x + 7 to find the y-cooridnate of the vertex:
y = 9 -18 + 7 = -2

so we know the vertex is the point (-3, -2)

alternatively we can complete the square:
y = x^2 + 6x+7 = x^2 + 6x + 9 - 2 = (x+3)^2 -2

we know that a parabola in the form y = (x-h)^2 +k has a vertex of (h, k)
in this case h = -3, k = -2, so we know the vertex is the point (-3, -2)

 

[Drongoski]

The focal point of a parabola is not its vertex.

I think the focal point for this parabola is: S(-3,-1.75)

 

[liamkk112]

The focal point of a parabola is not its vertex.

oops, this is not in the hsc syllabus though right?

 

[Luukas.2]

oops, this is not in the hsc syllabus though right?

Not any more, but it was until the start of 2020.

A parabola has its vertex at and a focal length of (which is the vertical distance from the vertex to the focus).

So, for example, has its vertex at and a focal length of . As it is concave up, its focus is at .

And, has its vertex at and a focal length of . As it is concave down, its focus is at .

Now, for the equation given:

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matching the answer provided above by @Drongoski.